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Minimum length subarray containing all unique elements after Q operations

Last Updated : 12 Jul, 2025
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Given an array of size N containing all elements as 0 initially, and a Q queries containing range in the form of [L, R]. The task is to modify the array by adding 1 to each element in the range [L, R] for Q queries and then print the size of minimum length subarray containing all unique elements.
Note: 1-based indexing is used in the range [L, R].
Examples:

Input: N = 6, Q[4][] = {{1, 3}, {4, 6}, {3, 4}, {3, 3}} 
Output: 3 
Explanation: 
Initial array: arr[] = { 0, 0, 0, 0, 0, 0 } 
Query 1: updated arr[] = { 1, 1, 1, 0, 0, 0 }. 
Query 2: updated arr[] = { 1, 1, 1, 1, 1, 1 }. 
Query 3: updated arr[] = { 1, 1, 2, 2, 1, 1 }. 
Query 4: updated arr[] = { 1, 1, 3, 2, 1, 1 }. 
The subarray { 1, 3, 2 } is minimum subarray which contains all unique elements. Thus, the answer is 3.
Input: N = 8, Q[6][] = {{1, 4}, {3, 4}, {4, 5}, {5, 5}, {7, 8}, {8, 8}} 
Output: 4 
Explanation: 
After processing all queries, the array becomes = { 1, 1, 2, 3, 2, 0, 1, 2 }. 
The subarray { 3, 2, 0, 1 } is minimum subarray which contains all unique elements. Thus, the answer is 4.

Approach: The idea is to use the concept of Prefix sum array and Two pointers approach to this problem. 

  • Final Array after the queries can be computed by incrementing the value at the array by 1 at the index L and decrementing the value by 1 at the index R + 1.
Processing of a Query:
arr[L] += 1
arr[R + 1] -= 1

Below is the implementation of the above approach:

C++ 
// C++ implementation to find the
// minimum size subarray containing
// all unique elements after processing
// the array for K queries of ranges

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

// Function to find minimum size subarray
// of all array elements
int subarrayLength(int A[], int R[][2],
                   int N, int M)
{

    // Updating the array after
    // processing each query
    for (int i = 0; i < M; ++i) {

        int l = R[i][0], r = R[i][1] + 1;

        // Making it to 0-indexing
        l--;
        r--;

        // Prefix sum array concept is used
        // to obtain the count array
        A[l]++;

        if (r < N)
            A[r]--;
    }

    // Iterating over the array
    // to get the final array
    for (int i = 1; i < N; ++i) {
        A[i] += A[i - 1];
    }

    // Variable to get count
    // of all unique elements
    int count = 0;

    // Hash to maintain previously
    // occurred elements
    unordered_set<int> s;

    // Loop to find the all
    // unique elements
    for (int i = 0; i < N; ++i) {
        if (s.find(A[i]) == s.end())
            count++;

        s.insert(A[i]);
    }

    // array to maintain counter
    // of encountered elements
    vector<int> repeat(count + 1, 0);

    // variable to store answer
    int ans = N;

    // Using two pointers approach
    int counter = 0, left = 0, right = 0;

    while (right < N) {

        int cur_element = A[right];
        repeat[cur_element] += 1;

        // Increment counter
        // if occurred once
        if (repeat[cur_element] == 1)
            ++counter;

        // when all unique
        // elements are found
        while (counter == count) {

            // update answer with
            // minimum size
            ans = min(ans, right - left + 1);

            // decrement count of
            // elements from left
            cur_element = A[left];
            repeat[cur_element] -= 1;
            ++left;

            // decrement counter
            if (repeat[cur_element] == 0)
                --counter;
        }

        ++right;
    }
    return ans;
}

// Driver code
int main()
{
    int N = 8, queries = 6;
    int Q[][2]
        = {
            { 1, 4 }, { 3, 4 }, { 4, 5 },
            { 5, 5 }, { 7, 8 }, { 8, 8 }
          };

    int A[N] = { 0 };

    cout << subarrayLength(A, Q, N, queries);

    return 0;
}
Java 
// Java implementation to find the
// minimum size subarray containing
// all unique elements after processing
// the array for K queries of ranges
import java.util.*;
class GFG{

// Function to find minimum size subarray
// of all array elements
static int subarrayLength(int A[], int R[][],
                          int N, int M)
{
    // Updating the array after
    // processing each query
    for (int i = 0; i < M; ++i) 
    {
        int l = R[i][0], r = R[i][1] + 1;

        // Making it to 0-indexing
        l--;
        r--;

        // Prefix sum array concept is used
        // to obtain the count array
        A[l]++;

        if (r < N)
            A[r]--;
    }

    // Iterating over the array
    // to get the final array
    for (int i = 1; i < N; ++i) 
    {
        A[i] += A[i - 1];
    }

    // Variable to get count
    // of all unique elements
    int count = 0;

    // Hash to maintain previously
    // occurred elements
    HashSet<Integer> s = new HashSet<Integer>();

    // Loop to find the all
    // unique elements
    for (int i = 0; i < N; ++i) 
    {
        if (!s.contains(A[i]))
            count++;
        s.add(A[i]);
    }

    // array to maintain counter
    // of encountered elements
    int []repeat = new int[count + 1];

    // variable to store answer
    int ans = N;

    // Using two pointers approach
    int counter = 0, left = 0, right = 0;

    while (right < N) 
    {
        int cur_element = A[right];
        repeat[cur_element] += 1;

        // Increment counter
        // if occurred once
        if (repeat[cur_element] == 1)
            ++counter;

        // when all unique
        // elements are found
        while (counter == count) 
        {
            // update answer with
            // minimum size
            ans = Math.min(ans, 
                           right - left + 1);

            // decrement count of
            // elements from left
            cur_element = A[left];
            repeat[cur_element] -= 1;
            ++left;

            // decrement counter
            if (repeat[cur_element] == 0)
                --counter;
        }

        ++right;
    }
    return ans;
}

// Driver code
public static void main(String[] args)
{
    int N = 8, queries = 6;
    int Q[][] = {{ 1, 4 }, { 3, 4 }, { 4, 5 },
                 { 5, 5 }, { 7, 8 }, { 8, 8 }};
    int A[] = new int[N];
    System.out.print(subarrayLength(A, Q, 
                                    N, queries));
}
}

// This code is contributed by Rajput-Ji
Python3 
# Python3 implementation to find the
# minimum size subarray containing
# all unique elements after processing
# the array for K queries of ranges

# Function to find minimum size subarray
# of all array elements
def subarrayLength(A, R, N, M):

    # Updating the array after
    # processing each query
    for i in range(M):

        l = R[i][0]
        r = R[i][1] + 1

        # Making it to 0-indexing
        l -= 1
        r -= 1

        # Prefix sum array concept is used
        # to obtain the count array
        A[l] += 1

        if (r < N):
            A[r] -= 1

    # Iterating over the array
    # to get the final array
    for i in range(1 , N):
        A[i] += A[i - 1]

    # Variable to get count
    # of all unique elements
    count = 0

    # Hash to maintain previously
    # occurred elements
    s = []

    # Loop to find the all
    # unique elements
    for i in range(N):
        if (A[i] not in s):
            count += 1

        s.append(A[i])

    # Array to maintain counter
    # of encountered elements
    repeat = [0] * (count + 1)

    # Variable to store answer
    ans = N

    # Using two pointers approach
    counter, left, right = 0, 0, 0

    while (right < N):

        cur_element = A[right]
        repeat[cur_element] += 1

        # Increment counter
        # if occurred once
        if (repeat[cur_element] == 1):
            counter += 1

        # When all unique
        # elements are found
        while (counter == count):

            # update answer with
            # minimum size
            ans = min(ans, right - left + 1)

            # Decrement count of
            # elements from left
            cur_element = A[left]
            repeat[cur_element] -= 1
            left += 1

            # Decrement counter
            if (repeat[cur_element] == 0):
                counter -= 1
                
        right += 1
        
    return ans

# Driver code
if __name__ == "__main__":
    
    N , queries = 8 , 6
    Q = [ [ 1, 4 ], [ 3, 4 ], [ 4, 5 ],
          [ 5, 5 ], [ 7, 8 ], [ 8, 8 ] ]

    A = [0] * N
    print(subarrayLength(A, Q, N, queries))

# This code is contributed by chitranayal
C# 
// C# implementation to find the
// minimum size subarray containing
// all unique elements after processing
// the array for K queries of ranges
using System;
using System.Collections.Generic;

class GFG{

// Function to find minimum size subarray
// of all array elements
static int subarrayLength(int []A, int [,]R,
                          int N, int M)
{
    
    // Updating the array after
    // processing each query
    for(int i = 0; i < M; ++i) 
    {
        int l = R[i, 0], r = R[i, 1] + 1;

        // Making it to 0-indexing
        l--;
        r--;

        // Prefix sum array concept is used
        // to obtain the count array
        A[l]++;

        if (r < N)
            A[r]--;
    }

    // Iterating over the array
    // to get the readonly array
    for(int i = 1; i < N; ++i) 
    {
        A[i] += A[i - 1];
    }

    // Variable to get count
    // of all unique elements
    int count = 0;

    // Hash to maintain previously
    // occurred elements
    HashSet<int> s = new HashSet<int>();

    // Loop to find the all
    // unique elements
    for(int i = 0; i < N; ++i) 
    {
        if (!s.Contains(A[i]))
            count++;
            
        s.Add(A[i]);
    }

    // Array to maintain counter
    // of encountered elements
    int []repeat = new int[count + 1];

    // Variable to store answer
    int ans = N;

    // Using two pointers approach
    int counter = 0, left = 0, right = 0;

    while (right < N) 
    {
        int cur_element = A[right];
        repeat[cur_element] += 1;

        // Increment counter
        // if occurred once
        if (repeat[cur_element] == 1)
            ++counter;

        // When all unique
        // elements are found
        while (counter == count) 
        {
            
            // Update answer with
            // minimum size
            ans = Math.Min(ans, 
                           right - left + 1);

            // Decrement count of
            // elements from left
            cur_element = A[left];
            repeat[cur_element] -= 1;
            ++left;

            // Decrement counter
            if (repeat[cur_element] == 0)
                --counter;
        }
        ++right;
    }
    return ans;
}

// Driver code
public static void Main(String[] args)
{
    int N = 8, queries = 6;
    int [,]Q = { { 1, 4 }, { 3, 4 }, { 4, 5 },
                 { 5, 5 }, { 7, 8 }, { 8, 8 } };
    int []A = new int[N];
    
    Console.Write(subarrayLength(A, Q, 
                                 N, queries));
}
}

// This code is contributed by Amit Katiyar
JavaScript 
<script>

// Javascript implementation to find the
// minimum size subarray containing
// all unique elements after processing
// the array for K queries of ranges

// Function to find minimum size subarray
// of all array elements
function subarrayLength(A, R, N, M)
{
    // Updating the array after
    // processing each query
    for (let i = 0; i < M; ++i) 
    {
        let l = R[i][0], r = R[i][1] + 1;
  
        // Making it to 0-indexing
        l--;
        r--;
  
        // Prefix sum array concept is used
        // to obtain the count array
        A[l]++;
  
        if (r < N)
            A[r]--;
    }
  
    // Iterating over the array
    // to get the final array
    for (let i = 1; i < N; ++i) 
    {
        A[i] += A[i - 1];
    }
  
    // Variable to get count
    // of all unique elements
    let count = 0;
  
    // Hash to maintain previously
    // occurred elements
    let s = new Set();
  
    // Loop to find the all
    // unique elements
    for (let i = 0; i < N; ++i) 
    {
        if (!s.has(A[i]))
            count++;
        s.add(A[i]);
    }
  
    // array to maintain counter
    // of encountered elements
    let repeat = Array.from({length: count + 1}, (_, i) => 0); 
  
    // variable to store answer
    let ans = N;
  
    // Using two pointers approach
    let counter = 0, left = 0, right = 0;
  
    while (right < N) 
    {
        let cur_element = A[right];
        repeat[cur_element] += 1;
  
        // Increment counter
        // if occurred once
        if (repeat[cur_element] == 1)
            ++counter;
  
        // when all unique
        // elements are found
        while (counter == count) 
        {
            // update answer with
            // minimum size
            ans = Math.min(ans, 
                           right - left + 1);
  
            // decrement count of
            // elements from left
            cur_element = A[left];
            repeat[cur_element] -= 1;
            ++left;
  
            // decrement counter
            if (repeat[cur_element] == 0)
                --counter;
        }
  
        ++right;
    }
    return ans;
}

// Driver code
    
      let N = 8, queries = 6;
    let Q = [[ 1, 4 ], [ 3, 4 ], [ 4, 5 ],
                 [ 5, 5 ], [ 7, 8 ], [ 8, 8 ]];
    let A = Array.from({length: N}, (_, i) => 0);
    document.write(subarrayLength(A, Q, 
                                    N, queries));
                                                                                     
</script>

Output: 
4

Time Complexity: O(N) 


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