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Smallest Range with Elements from k Sorted Lists

Last Updated : 15 May, 2025
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Given a 2d integer array arr[][] of order k * n, where each row is sorted in ascending order. Your task is to find the smallest range that includes at least one element from each of the K lists. If more than one such ranges are found, return the first one.

Examples: 

Input: arr[][] = [[ 4, 7, 9, 12, 15 ],
[ 0, 8, 10, 14, 20 ],
[ 6, 12, 16, 30, 50 ]]
Output: 6 8
Explanation: Smallest range is formed by  number 7 from the first list, 8 from second list and 6 from the third list.

Input: arr[][] = [[ 2, 4 ],
[ 1, 7 ],
[ 20, 40 ]]
Output: 4 20
Explanation: The range [4, 20] contains 4, 7, 20 which contains element from all the three arrays.

[Naive Approach] - Using K Pointers - O(n k^2) Time and O(k) Space

The idea is to keep k pointers one for each list, starting at index 0. At each step, take the min and max of the current K elements to form a range. To minimize the range, we must increase the min value, since we can't decrease the max (all pointers start at 0). So, move the pointer of the list that has the current minimum, and update the range. Repeat until one list is exhausted.

Step By Step Implementation:

  • Create a list of pointers, one for each input list, all starting at index 0.
  • Repeat the process until one of the pointers reaches the end of its list.
  • At each step, pick the current elements pointed by all pointers.
  • Find the minimum and maximum among those elements.
  • Calculate the range using the min and max values.
  • If this range is smaller than the previous best, update the answer.
  • Move forward the pointer of the list that had the minimum element.
  • Stop when any list is exhausted, and return the best range found.
C++
// C++ program to find the smallest range
// that includes at least one element from
// each of the k sorted lists using k pointers
#include <iostream>
#include <vector>
#include <climits>
using namespace std;

vector<int> findSmallestRange(vector<vector<int>>& arr) {
    
    int k = arr.size();         
    int n = arr[0].size();   

    // Pointers for each of the k rows
    vector<int> ptr(k, 0);

    int minRange = INT_MAX;
    int start = -1, end = -1;

    while (true) {

        int minVal = INT_MAX;
        int maxVal = INT_MIN;
        int minRow = -1;

        // Traverse all k rows to get current min and max
        for (int i = 0; i < k; i++) {

            // If any list is exhausted, stop the loop
            if (ptr[i] == n) {
                return {start, end};
            }

            // Track min value and its row index
            if (arr[i][ptr[i]] < minVal) {
                minVal = arr[i][ptr[i]];
                minRow = i;
            }

            // Track current max value
            if (arr[i][ptr[i]] > maxVal) {
                maxVal = arr[i][ptr[i]];
            }
        }

        // Update the result range if a 
        // smaller range is found
        if (maxVal - minVal < minRange) {
            minRange = maxVal - minVal;
            start = minVal;
            end = maxVal;
        }

        // Move the pointer of the 
        // row with minimum value
        ptr[minRow]++;
    }

    return {start, end};
}

int main() {

    vector<vector<int>> arr = {
        {4, 7, 9, 12, 15},
        {0, 8, 10, 14, 20},
        {6, 12, 16, 30, 50}
    };

    vector<int> res = findSmallestRange(arr);

    cout << res[0] << " " << res[1];

    return 0;
}
Java
// Java program to find the smallest range
import java.util.*;

class GfG{
    static ArrayList<Integer> findSmallestRange(int[][] arr) {
        int k = arr.length;
        int n = arr[0].length;

        // Pointers for each of the k rows
        int[] ptr = new int[k];

        int minRange = Integer.MAX_VALUE;
        int start = -1, end = -1;

        while (true) {
            int minVal = Integer.MAX_VALUE;
            int maxVal = Integer.MIN_VALUE;
            int minRow = -1;

            // Traverse all k rows to get current min and max
            for (int i = 0; i < k; i++) {
                // If any list is exhausted, stop the loop
                if (ptr[i] == n) {
                    ArrayList<Integer> result = new ArrayList<>();
                    result.add(start);
                    result.add(end);
                    return result;
                }

                // Track min value and its row index
                if (arr[i][ptr[i]] < minVal) {
                    minVal = arr[i][ptr[i]];
                    minRow = i;
                }

                // Track current max value
                if (arr[i][ptr[i]] > maxVal) {
                    maxVal = arr[i][ptr[i]];
                }
            }

            // Update the result range if a smaller range is found
            if (maxVal - minVal < minRange) {
                minRange = maxVal - minVal;
                start = minVal;
                end = maxVal;
            }

            // Move the pointer of the row with minimum value
            ptr[minRow]++;
        }
    }

    public static void main(String[] args) {
        int[][] arr = {
            {4, 7, 9, 12, 15},
            {0, 8, 10, 14, 20},
            {6, 12, 16, 30, 50}
        };

        ArrayList<Integer> res = findSmallestRange(arr);

        System.out.println(res.get(0) + " " + res.get(1));
    }
}
Python
# Python program to find the smallest range

def findSmallestRange(arr):
    k = len(arr)
    n = len(arr[0])

    # Pointers for each of the k rows
    ptr = [0] * k

    min_range = float('inf')
    start = -1
    end = -1

    while True:
        min_val = float('inf')
        max_val = float('-inf')
        min_row = -1

        # Traverse all k rows to get current min and max
        for i in range(k):
            # If any list is exhausted, stop the loop
            if ptr[i] == n:
                return [start, end]

            # Track min value and its row index
            if arr[i][ptr[i]] < min_val:
                min_val = arr[i][ptr[i]]
                min_row = i

            # Track current max value
            if arr[i][ptr[i]] > max_val:
                max_val = arr[i][ptr[i]]

        # Update the result range if a smaller range is found
        if max_val - min_val < min_range:
            min_range = max_val - min_val
            start = min_val
            end = max_val

        # Move the pointer of the row with minimum value
        ptr[min_row] += 1


if __name__ == "__main__":
    arr = [
        [4, 7, 9, 12, 15],
        [0, 8, 10, 14, 20],
        [6, 12, 16, 30, 50]
    ]
    
    res = findSmallestRange(arr)
    
    print(res[0], res[1])
C#
using System;
using System.Collections.Generic;


class GfG{
    static List<int> findSmallestRange(int[,] arr) {
        int k = arr.GetLength(0);
        int n = arr.GetLength(1);
        // Pointers for each of the k rows
        int[] ptr = new int[k]; 

        int minRange = int.MaxValue;
        int start = -1, end = -1;

        while (true) {
            int minVal = int.MaxValue;
            int maxVal = int.MinValue;
            int minRow = -1;

            // Traverse all k rows to get current min and max
            for (int i = 0; i < k; i++) {
                // If any list is exhausted, stop the loop
                if (ptr[i] == n) {
                    return new List<int> { start, end };
                }

                int current = arr[i, ptr[i]];

                if (current < minVal) {
                    minVal = current;
                    minRow = i;
                }

                if (current > maxVal) {
                    maxVal = current;
                }
            }

            // Update the result range if a smaller range is found
            if (maxVal - minVal < minRange) {
                minRange = maxVal - minVal;
                start = minVal;
                end = maxVal;
            }

            // Move the pointer of the row with minimum value
            ptr[minRow]++;
        }
    }

    public static void Main(string[] args) {
        int[,] arr = {
            { 4, 7, 9, 12, 15 },
            { 0, 8, 10, 14, 20 },
            { 6, 12, 16, 30, 50 }
        };

        List<int> res = findSmallestRange(arr);

        Console.WriteLine(res[0] + " " + res[1]);
    }
}
JavaScript
// JavaScript program to find the smallest range
function findSmallestRange(arr) {
    let k = arr.length;
    let n = arr[0].length;

    // Pointers for each of the k rows
    let ptr = new Array(k).fill(0);

    let minRange = Infinity;
    let start = -1, end = -1;

    while (true) {
        let minVal = Infinity;
        let maxVal = -Infinity;
        let minRow = -1;

        // Traverse all k rows to get current min and max
        for (let i = 0; i < k; i++) {
            // If any list is exhausted, stop the loop
            if (ptr[i] === n) {
                return [start, end];
            }

            // Track min value and its row index
            if (arr[i][ptr[i]] < minVal) {
                minVal = arr[i][ptr[i]];
                minRow = i;
            }

            // Track current max value
            if (arr[i][ptr[i]] > maxVal) {
                maxVal = arr[i][ptr[i]];
            }
        }

        // Update the result range if a smaller range is found
        if (maxVal - minVal < minRange) {
            minRange = maxVal - minVal;
            start = minVal;
            end = maxVal;
        }

        // Move the pointer of the row with minimum value
        ptr[minRow]++;
    }
}

const arr = [
    [4, 7, 9, 12, 15],
    [0, 8, 10, 14, 20],
    [6, 12, 16, 30, 50]
];

const res = findSmallestRange(arr);

console.log(res[0] + ' ' + res[1]);

Output
6 8

[Better Approach] Using Two Pointer - O(n*k log (n*k)) Time and O(n*k) Space

The Idea is to find the smallest range problem by transforming it into a sliding window problem over a merged and sorted list of all elements from the input lists. Each element is stored along with its original list index to track its source. After sorting the combined list by value, two pointers (left and right) are used to define a window that moves through the list. As the window expands, a frequency map tracks how many unique lists are represented. When the window includes at least one number from every list, the algorithm tries to shrink it from the left to find a smaller valid range. The smallest such range found during this process is returned as the result.

C++
#include <bits/stdc++.h>
using namespace std;

vector<int> findSmallestRange(vector<vector<int>>& arr) {
    
    int k = arr.size();  
    // Stores the current index for each list
    vector<int> pointers(k, 0);
    // Stores the current smallest range
    vector<int> smallestRange = {0, INT_MAX};

    while (true) {
        int currentMin = INT_MAX, currentMax = INT_MIN;
        int minListIndex = -1;

        // Find the minimum and maximum among current elements of all lists
        for (int i = 0; i < k; i++) {
            int value = arr[i][pointers[i]];

            if (value < currentMin) {
                currentMin = value;
                minListIndex = i;
            }

            if (value > currentMax) {
                currentMax = value;
            }
        }

        // Update the smallest range if this one is smaller
        if (currentMax - currentMin < smallestRange[1] - smallestRange[0]) {
            smallestRange[0] = currentMin;
            smallestRange[1] = currentMax;
        }

        // Move the pointer in the list that had the minimum value
        pointers[minListIndex]++;
        // If that list is exhausted, break the loop
        if (pointers[minListIndex] == arr[minListIndex].size()) break;
    }

    return smallestRange;
}

// Driver code
int main() {
    vector<vector<int>> arr = {
        {4, 7, 9, 12, 15},
        {0, 8, 10, 14, 20},
        {6, 12, 16, 30, 50}
    };

    vector<int> result = findSmallestRange(arr);
    cout << result[0] << " " << result[1];

    return 0;
}
Java
import java.util.*;

class GfG {

    // Function to find the smallest range
    public static ArrayList<Integer> findSmallestRange(int[][] arr) {
        int k = arr.length; // Number of lists
        // Stores the current index for each list
        int[] pointers = new int[k];
        // Stores the current smallest range
        ArrayList<Integer> smallestRange = new ArrayList<>
                                   (Arrays.asList(0, Integer.MAX_VALUE));

        // Continue the loop until one list is exhausted
        while (true) {
            int currentMin = Integer.MAX_VALUE, currentMax = Integer.MIN_VALUE;
            int minListIndex = -1;

            // Find the minimum and maximum among current elements of all lists
            for (int i = 0; i < k; i++) {
                int value = arr[i][pointers[i]];

                // Update the current minimum
                if (value < currentMin) {
                    currentMin = value;
                    minListIndex = i;
                }

                // Update the current maximum
                if (value > currentMax) {
                    currentMax = value;
                }
            }

            // Update the smallest range if this one is smaller
            if (currentMax - currentMin < smallestRange.get(1) - smallestRange.get(0)) {
                smallestRange.set(0, currentMin);
                smallestRange.set(1, currentMax);
            }

            // Move the pointer in the list that had the minimum value
            pointers[minListIndex]++;
            // If that list is exhausted, break the loop
            if (pointers[minListIndex] == arr[minListIndex].length) break;
        }

        return smallestRange; // Return the result as ArrayList
    }

    // Driver code
    public static void main(String[] args) {
        int[][] arr = {
            {4, 7, 9, 12, 15},
            {0, 8, 10, 14, 20},
            {6, 12, 16, 30, 50}
        };

        ArrayList<Integer> result = findSmallestRange(arr);
        System.out.println(result.get(0) + " " + result.get(1));
    }
}
Python
def findSmallestRange(arr):
    k = len(arr)  # Number of lists
    # Stores the current index for each list
    pointers = [0] * k
    # Stores the current smallest range
    smallestRange = [0, float('inf')]

    # Continue the loop until one list is exhausted
    while True:
        currentMin = float('inf')
        currentMax = -float('inf')
        minListIndex = -1

        # Find the minimum and maximum among current elements of all lists
        for i in range(k):
            value = arr[i][pointers[i]]

            # Update the current minimum
            if value < currentMin:
                currentMin = value
                minListIndex = i

            # Update the current maximum
            if value > currentMax:
                currentMax = value

        # Update the smallest range if this one is smaller
        if currentMax - currentMin < smallestRange[1] - smallestRange[0]:
            smallestRange[0] = currentMin
            smallestRange[1] = currentMax

        # Move the pointer in the list that had the minimum value
        pointers[minListIndex] += 1
        # If that list is exhausted, break the loop
        if pointers[minListIndex] == len(arr[minListIndex]):
            break

    return smallestRange  # Return the result as a list

# Driver code
if __name__ == "__main__":
    arr = [
        [4, 7, 9, 12, 15],
        [0, 8, 10, 14, 20],
        [6, 12, 16, 30, 50]
    ]
    
    result = findSmallestRange(arr)
    print(result[0], result[1])
C#
using System;
using System.Collections.Generic;

class GfG{

    // Function to find the smallest range
    public static List<int> findSmallestRange(int[,] arr) {
        int k = arr.GetLength(0); // Number of lists (rows)
        // Stores the current index for each list (row)
        int[] pointers = new int[k];
        // Stores the current smallest range
        List<int> smallestRange = new List<int> { 0, int.MaxValue };

        // Continue the loop until one list is exhausted
        while (true) {
            int currentMin = int.MaxValue, currentMax = int.MinValue;
            int minListIndex = -1;

            // Find the minimum and maximum among current elements 
            // of all lists
            for (int i = 0; i < k; i++) {
                int value = arr[i, pointers[i]];

                // Update the current minimum
                if (value < currentMin) {
                    currentMin = value;
                    minListIndex = i;
                }

                // Update the current maximum
                if (value > currentMax) {
                    currentMax = value;
                }
            }

            // Update the smallest range if this one is smaller
            if (currentMax - currentMin < smallestRange[1] - smallestRange[0]) {
                smallestRange[0] = currentMin;
                smallestRange[1] = currentMax;
            }

            // Move the pointer in the list that had the minimum value
            pointers[minListIndex]++;
            // If that list is exhausted, break the loop
            if (pointers[minListIndex] == arr.GetLength(1)) break;
        }

        return smallestRange; // Return the result as List<int>
    }

    // Driver code
    public static void Main(string[] args) {
        int[,] arr = {
            {4, 7, 9, 12, 15},
            {0, 8, 10, 14, 20},
            {6, 12, 16, 30, 50}
        };

        List<int> result = findSmallestRange(arr);
        Console.WriteLine(result[0] + " " + result[1]);
    }
}
JavaScript
function findSmallestRange(arr) {
    const k = arr.length;  // Number of lists
    // Stores the current index for each list
    let pointers = new Array(k).fill(0);
    // Stores the current smallest range
    let smallestRange = [0, Number.MAX_VALUE];

    // Continue the loop until one list is exhausted
    while (true) {
        let currentMin = Number.MAX_VALUE, currentMax = -Number.MAX_VALUE;
        let minListIndex = -1;

        // Find the minimum and maximum among current elements of all lists
        for (let i = 0; i < k; i++) {
            const value = arr[i][pointers[i]];

            // Update the current minimum
            if (value < currentMin) {
                currentMin = value;
                minListIndex = i;
            }

            // Update the current maximum
            if (value > currentMax) {
                currentMax = value;
            }
        }

        // Update the smallest range if this one is smaller
        if (currentMax - currentMin < smallestRange[1] - smallestRange[0]) {
            smallestRange[0] = currentMin;
            smallestRange[1] = currentMax;
        }

        // Move the pointer in the list that had the minimum value
        pointers[minListIndex]++;
        // If that list is exhausted, break the loop
        if (pointers[minListIndex] === arr[minListIndex].length) break;
    }

    return smallestRange;  // Return the result as an array
}

// Driver code
const arr = [
    [4, 7, 9, 12, 15],
    [0, 8, 10, 14, 20],
    [6, 12, 16, 30, 50]
];

const result = findSmallestRange(arr);
console.log(result[0], result[1]);

Output
6 8

[Efficient Approach] - Using Min Heap - O(n k Log k) Time and O(k) Space

Min-Heap can be used to find the minimum value in logarithmic time or log k time instead of linear time. To find the max value, we initialize the max value of all 0 indexes initially. For rest of the max values in the loop, we simply compare the current max value with the next item from the list from which the min item is being removed. Rest of the approach remains the same. 

Step By Step Implementation:

  • Create a Min-Heap to store K elements, one from each array, and a variable minrange initialized to a maximum value and also keep a variable max to store the maximum integer.
  • Initially put the first element from each list and store the maximum value in max.
  • Repeat the following steps until at least one list exhausts : 
    • find the minimum value or min, use the top or root of the Min heap which is the minimum element.
    • Now update the minrange if the current (max-min) is less than minrange.
    • Remove the top or root element from the priority queue, insert the next element from the list containing the min element
    • Update the max with the new element inserted if the new element is greater than the previous max.
C++
#include <bits/stdc++.h>
using namespace std;

// Struct to represent elements in the heap
struct Node {
    int val, row, col;
    bool operator>(const Node& other) const {
        return val > other.val;
    }
};

// Function to find the smallest range
vector<int> findSmallestRange(vector<vector<int>>& arr) {
    int N = arr.size();        // Number of rows
    int K = arr[0].size();     // Number of columns (same for each row)
    priority_queue<Node, vector<Node>, greater<Node>> pq;
    int maxVal = INT_MIN;

    // Push the first element of each list into the min-heap
    for (int i = 0; i < N; i++) {
        pq.push({arr[i][0], i, 0});
        maxVal = max(maxVal, arr[i][0]);
    }

    int minRange = INT_MAX, minEl, maxEl;

    while (true) {
        Node curr = pq.top(); pq.pop();
        int minVal = curr.val;

        // Update range if better
        if (maxVal - minVal < minRange) {
            minRange = maxVal - minVal;
            minEl = minVal;
            maxEl = maxVal;
        }

        // If we've reached the end of a list, break
        if (curr.col + 1 == K) break;

        // Push next element from the same list
        int nextVal = arr[curr.row][curr.col + 1];
        pq.push({nextVal, curr.row, curr.col + 1});
        maxVal = max(maxVal, nextVal);
    }

    return {minEl, maxEl};
}

// Driver code
int main() {
    vector<vector<int>> arr = {
        {4, 7, 9, 12, 15},
        {0, 8, 10, 14, 20},
        {6, 12, 16, 30, 50}
    };

    vector<int> result = findSmallestRange(arr);
    cout << result[0] << " " << result[1];

    return 0;
}
Java
import java.util.*;

// Class to represent elements in the heap
class Node implements Comparable<Node> {
    int val, row, col;

    Node(int val, int row, int col) {
        this.val = val;
        this.row = row;
        this.col = col;
    }

    // For min-heap based on value
    public int compareTo(Node other) {
        return this.val - other.val;
    }
}

class GfG {
    // Function to find the smallest range
    static ArrayList<Integer> findSmallestRange(int[][] arr) {
        int k = arr.length;
        int n = arr[0].length;
        PriorityQueue<Node> pq = new PriorityQueue<>();
        int maxVal = Integer.MIN_VALUE;

        // Push the first element of each list into the min-heap
        for (int i = 0; i < k; i++) {
            pq.add(new Node(arr[i][0], i, 0));
            maxVal = Math.max(maxVal, arr[i][0]);
        }

        int minRange = Integer.MAX_VALUE, minEl = -1, maxEl = -1;

        while (true) {
            Node curr = pq.poll();
            int minVal = curr.val;

            // Update range if better
            if (maxVal - minVal < minRange) {
                minRange = maxVal - minVal;
                minEl = minVal;
                maxEl = maxVal;
            }

            // If we've reached the end of a list, break
            if (curr.col + 1 == n)
                break;

            // Push next element from the same list
            int nextVal = arr[curr.row][curr.col + 1];
            pq.add(new Node(nextVal, curr.row, curr.col + 1));
            maxVal = Math.max(maxVal, nextVal);
        }

        // Return result as ArrayList
        ArrayList<Integer> result = new ArrayList<>();
        result.add(minEl);
        result.add(maxEl);
        return result;
    }

    // Driver code
    public static void main(String[] args) {
        int[][] arr = {
            {4, 7, 9, 12, 15},
            {0, 8, 10, 14, 20},
            {6, 12, 16, 30, 50}
        };

        ArrayList<Integer> res = findSmallestRange(arr);
        System.out.println(res.get(0) + " " + res.get(1));
    }
}
Python
import heapq

# Function to find the smallest range
def findSmallestRange(arr):
    k = len(arr)
    n = len(arr[0])
    heap = []
    maxVal = float('-inf')

    # Push the first element of each 
    # list into the min-heap
    for i in range(k):
        heapq.heappush(heap, (arr[i][0], i, 0))
        maxVal = max(maxVal, arr[i][0])

    minRange = float('inf')
    minEl = maxEl = -1

    while True:
        minVal, row, col = heapq.heappop(heap)

        # Update range if better
        if maxVal - minVal < minRange:
            minRange = maxVal - minVal
            minEl = minVal
            maxEl = maxVal

        # If we've reached the end of a list, break
        if col + 1 == n:
            break

        # Push next element from the same list
        nextVal = arr[row][col + 1]
        heapq.heappush(heap, (nextVal, row, col + 1))
        maxVal = max(maxVal, nextVal)

    return [minEl, maxEl]

# Driver code
if __name__ == "__main__":
    arr = [
        [4, 7, 9, 12, 15],
        [0, 8, 10, 14, 20],
        [6, 12, 16, 30, 50]
    ]
    
    res = findSmallestRange(arr)
    print(res[0], res[1])
C#
using System;
using System.Collections.Generic;

// Class to represent elements in the heap
class Node : IComparable<Node> {
    public int val, row, col;

    public Node(int val, int row, int col) {
        this.val = val;
        this.row = row;
        this.col = col;
    }

    // For min-heap based on value
    public int CompareTo(Node other) {
        if (this.val != other.val)
            return this.val.CompareTo(other.val);
        // To avoid duplicate keys in SortedSet
        if (this.row != other.row)
            return this.row.CompareTo(other.row);
        return this.col.CompareTo(other.col);
    }
}

class GfG {
    // Function to find the smallest range
    static List<int> findSmallestRange(int[,] arr) {
        int k = arr.GetLength(0);
        int n = arr.GetLength(1);
        var pq = new SortedSet<Node>();
        int maxVal = int.MinValue;

        // Push the first element of each list into the min-heap
        for (int i = 0; i < k; i++) {
            var node = new Node(arr[i, 0], i, 0);
            pq.Add(node);
            maxVal = Math.Max(maxVal, arr[i, 0]);
        }

        int minRange = int.MaxValue, minEl = -1, maxEl = -1;

        while (true) {
            var curr = GetMin(pq);
            pq.Remove(curr);
            int minVal = curr.val;

            // Update range if better
            if (maxVal - minVal < minRange) {
                minRange = maxVal - minVal;
                minEl = minVal;
                maxEl = maxVal;
            }

            // If we've reached the end of a list, break
            if (curr.col + 1 == n)
                break;

            // Push next element from the same list
            int nextVal = arr[curr.row, curr.col + 1];
            var nextNode = new Node(nextVal, curr.row, curr.col + 1);
            pq.Add(nextNode);
            maxVal = Math.Max(maxVal, nextVal);
        }

        return new List<int> { minEl, maxEl }; // Return result as List<int>
    }

    // Helper to get the minimum element (first element in SortedSet)
    static Node GetMin(SortedSet<Node> pq) {
        foreach (var node in pq)
            return node;
        return null;
    }

    // Driver code
    static void Main() {
        int[,] arr = {
            {4, 7, 9, 12, 15},
            {0, 8, 10, 14, 20},
            {6, 12, 16, 30, 50}
        };

        List<int> res = findSmallestRange(arr);
        Console.WriteLine(res[0] + " " + res[1]);
    }
}
JavaScript
class Node {
    constructor(val, row, col) {
        this.val = val;
        this.row = row;
        this.col = col;
    }
}

// Function to find the smallest range
function findSmallestRange(arr) {
    const k = arr.length;
    const n = arr[0].length;
    const heap = new MinHeap();
    let maxVal = -Infinity;

    // Push the first element of each list into the min-heap
    for (let i = 0; i < k; i++) {
        heap.push(new Node(arr[i][0], i, 0));
        maxVal = Math.max(maxVal, arr[i][0]);
    }

    let minRange = Infinity;
    let minEl = -1, maxEl = -1;

    while (true) {
        const curr = heap.pop();
        const minVal = curr.val;

        // Update range if better
        if (maxVal - minVal < minRange) {
            minRange = maxVal - minVal;
            minEl = minVal;
            maxEl = maxVal;
        }

        // If we've reached the end of a list, break
        if (curr.col + 1 === n) break;

        // Push next element from the same list
        const nextVal = arr[curr.row][curr.col + 1];
        heap.push(new Node(nextVal, curr.row, curr.col + 1));
        maxVal = Math.max(maxVal, nextVal);
    }

    return [minEl, maxEl];
}


// Min-heap comparator
class MinHeap {
    constructor() {
        this.heap = [];
    }

    push(node) {
        this.heap.push(node);
        this._heapifyUp();
    }

    pop() {
        if (this.size() === 1) return this.heap.pop();
        const top = this.heap[0];
        this.heap[0] = this.heap.pop();
        this._heapifyDown();
        return top;
    }

    top() {
        return this.heap[0];
    }

    size() {
        return this.heap.length;
    }

    _heapifyUp() {
        let idx = this.size() - 1;
        while (idx > 0) {
            let parent = Math.floor((idx - 1) / 2);
            if (this.heap[parent].val <= this.heap[idx].val) break;
            [this.heap[parent], this.heap[idx]] = [this.heap[idx], this.heap[parent]];
            idx = parent;
        }
    }

    _heapifyDown() {
        let idx = 0;
        const n = this.size();
        while (true) {
            let left = 2 * idx + 1;
            let right = 2 * idx + 2;
            let smallest = idx;

            if (left < n && this.heap[left].val < this.heap[smallest].val) {
                smallest = left;
            }
            if (right < n && this.heap[right].val < this.heap[smallest].val) {
                smallest = right;
            }
            if (smallest === idx) break;
            [this.heap[smallest], this.heap[idx]] = [this.heap[idx], this.heap[smallest]];
            idx = smallest;
        }
    }
}


// Driver code
const arr = [
    [4, 7, 9, 12, 15],
    [0, 8, 10, 14, 20],
    [6, 12, 16, 30, 50]
];

const res = findSmallestRange(arr);
console.log(res[0] + " " + res[1]);

Output
6 8



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