Find Subarray ranges having difference between max and min exactly K
Last Updated :
20 Jul, 2022
Given an array arr[] of length N and integer K, the task is to print subarray ranges (starting index, ending index) of the array where difference between max and min elements of the subarray is exactly K.( 1-based index )
Examples:
Input: arr[] = {2, 1, 3, 4, 2, 6}, K = 2
Output: (1, 3), (2, 3), (3, 5), (4, 5)
Explanation: In the above array following sub array ranges have max min difference exactly K
(1, 3) => max = 3 and min = 1. Difference = 3 - 1 = 2
(2, 3) => max = 3 and min = 1. Difference = 3 - 1 = 2
(3, 5) => max = 4 and min = 2. Difference = 4 - 2 = 2
(4, 5) => max = 4 and min = 2. Difference = 4 - 2 = 2
Input: arr[] = {5, 3, 4, 6, 1, 2}, K = 6
Output: -1
Explanation: There is no such sub array ranges.
Approach: The basic idea to solve the problem is to form all the subarrays and find the minimum and maximum and their difference for each subarray. Follow the steps mentioned below to solve the problem.
- Iterate over the array from i = 0 to N-1:
- Iterate from j = i to N-1:
- Insert arr[j] current element in a set storing the current subarray starting from i.
- Find the minimum and maximum of the set.
- Get their difference and check if their difference is equal to K or not.
- If it is, then push this range in the answer.
- Return the ranges.
- If no such range then return -1.
Below is the implementation of the above approach.
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print index ranges
void printRanges(vector<int> arr, int n,
int K)
{
int i, j, f = 0;
for (i = 0; i < n; i++) {
// Set to store the elements
// of a subarray
set<int> s;
for (j = i; j < n; j++) {
// Insert current element in set
s.insert(arr[j]);
// Calculate max and min for
// any particular index range
int max = *s.rbegin();
int min = *s.begin();
// If we get max-min = K
// print 1 based index
if (max - min == K) {
cout << i + 1 << " " << j + 1
<< "\n";
f = 1;
}
}
}
// If we didn't find any index ranges
if (f == 0)
cout << -1 << endl;
}
// Driver Code
int main()
{
vector<int> arr = { 2, 1, 3, 4, 2, 6 };
int N = arr.size();
int K = 2;
// Function call
printRanges(arr, N, K);
return 0;
}
// Java implementation of above approach
import java.util.*;
class GFG {
// Function to print index ranges
static void printRanges(int arr[], int n,
int K)
{
int i, j, f = 0;
for (i = 0; i < n; i++) {
// Set to store the elements
// of a subarray
Set<Integer> s = new HashSet<>();
for (j = i; j < n; j++) {
// Insert current element in set
s.add(arr[j]);
// Calculate max and min for
// any particular index range
int max = Collections.max(s);
int min = Collections.min(s);
// If we get max-min = K
// print 1 based index
if (max - min == K) {
System.out.println( (i + 1) + " " + (j + 1));
f = 1;
}
}
}
// If we didn't find any index ranges
if (f == 0)
System.out.println(-1);
}
// Driver Code
public static void main (String[] args) {
int arr[] = { 2, 1, 3, 4, 2, 6 };
int N = arr.length;
int K = 2;
// Function call
printRanges(arr, N, K);
}
}
// This code is contributed by hrithikgarg03188.
# python3 implementation of above approach
# Function to print index ranges
def printRanges(arr, n, K):
i, j, f = 0, 0, 0
for i in range(0, n):
# Set to store the elements
# of a subarray
s = set()
for j in range(i, n):
# Insert current element in set
s.add(arr[j])
# Calculate max and min for
# any particular index range
ma = max(list(s))
mi = min(list(s))
# If we get max-min = K
# print 1 based index
if (ma - mi == K):
print(f"{i + 1} {j + 1}")
f = 1
# If we didn't find any index ranges
if (f == 0):
print(-1)
# Driver Code
if __name__ == "__main__":
arr = [2, 1, 3, 4, 2, 6]
N = len(arr)
K = 2
# Function call
printRanges(arr, N, K)
# This code is contributed by rakeshsahni
// C# implementation of above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to print index ranges
static void printRanges(int[] arr, int n, int K)
{
int i, j, f = 0;
for (i = 0; i < n; i++) {
// Set to store the elements
// of a subarray
HashSet<int> s = new HashSet<int>();
for (j = i; j < n; j++) {
// Insert current element in set
s.Add(arr[j]);
// Calculate max and min for
// any particular index range
int max = int.MinValue,min = int.MaxValue;
foreach(var value in s)
{
max = Math.Max(max,value);
min = Math.Min(min,value);
}
// If we get max-min = K
// print 1 based index
if (max - min == K) {
Console.Write( (i + 1) + " " + (j + 1) + "\n");
f = 1;
}
}
}
// If we didn't find any index ranges
if (f == 0)
Console.Write(-1);
}
// Driver Code
public static void Main()
{
int[] arr = { 2, 1, 3, 4, 2, 6 };
int N = arr.Length;
int K = 2;
// Function call
printRanges(arr, N, K);
}
}
// This code is contributed by aditya942003patil.
<script>
// Javascript implementation of above approach
// Function to print index ranges
function printRanges(arr, n, K)
{
let i = 0, j = 0, f = 0;
for (i = 0; i < n; i++) {
// Set to store the elements
// of a subarray
const s = new Set();
for (j = i; j < n; j++) {
// Insert current element in set
s.add(arr[j]);
// Calculate max and min for
// any particular index range
let max = Math.max(...s);
let min = Math.min(...s);
// If we get max-min = K
// print 1 based index
if (max - min == K) {
document.write((i+1) + " " + (j+1) + "<br>");
f = 1;
}
}
}
// If we didn't find any index ranges
if (f == 0)
document.write(-1);
}
// Driver Code
let arr = [ 2, 1, 3, 4, 2, 6 ];
let N = arr.length;
let K = 2;
// Function call
printRanges(arr, N, K);
// This code is contributed by aditya942003patil.
</script>
Time Complexity: O(N2 * logN)
Auxiliary Space: O(N)
Efficient Approach: Instead of using the HashSet, keep a track of current maximum and minimum element while traversing the array.
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print index ranges
void printRanges(vector<int> arr, int n, int K)
{
int i = 0, j = 0, f = 0;
for (i = 0; i < n; i++) {
int mx = INT_MIN, mn = INT_MAX;
for (j = i; j < n; j++) {
// Calculate max and min for
// any particular index range
mx = max(mx, arr[j]);
mn = min(mn, arr[j]);
// If we get max-min = K
// print 1 based index
if (mx - mn == K) {
cout << (i + 1) << " " << (j + 1) << endl;
f = 1;
}
}
}
// If we didn't find any index ranges
if (f == 0)
cout << -1;
}
// Driver Code
int main()
{
vector<int> arr = { 2, 1, 3, 4, 2, 6 };
int N = arr.size();
int K = 2;
// Function call
printRanges(arr, N, K);
}
// This code is contributed by Samim Hossain Mondal.
// Java implementation of above approach
import java.util.*;
public class GFG {
// Function to print index ranges
static void printRanges(int[] arr, int n, int K)
{
int i = 0, j = 0, f = 0;
for (i = 0; i < n; i++) {
int mx = Integer.MIN_VALUE, mn
= Integer.MAX_VALUE;
for (j = i; j < n; j++) {
// Calculate max and min for
// any particular index range
mx = Math.max(mx, arr[j]);
mn = Math.min(mn, arr[j]);
// If we get max-min = K
// print 1 based index
if (mx - mn == K) {
System.out.println((i + 1) + " "
+ (j + 1));
f = 1;
}
}
}
// If we didn't find any index ranges
if (f == 0)
System.out.print(-1);
}
// Driver Code
public static void main(String args[])
{
int[] arr = { 2, 1, 3, 4, 2, 6 };
int N = arr.length;
int K = 2;
// Function call
printRanges(arr, N, K);
}
}
// This code is contributed by Samim Hossain Mondal.
# Python3 implementation of above approach
# import the module
import sys
# Function to print index ranges
def printRanges(arr, n, K):
i, j, f = 0, 0, 0
for i in range(0, n):
mn = sys.maxsize
mx = -1*sys.maxsize
for j in range(i, n):
# Calculate max and min for
# any particular index range
mx = max(mx, arr[j])
mn = min(mn, arr[j])
# If we get max-min=K
# print 1 based index
if(mx - mn == K):
print(f"{i + 1} {j + 1}")
f=1
# If we didn't find any index ranges
if (f == 0):
print(-1)
# Driver Code
if __name__ == "__main__":
arr = [2, 1, 3, 4, 2, 6]
N = len(arr)
K = 2
# Function call
printRanges(arr, N, K)
# This code is contributed by Pushpesh Raj
// C# implementation of above approach
using System;
class GFG {
// Function to print index ranges
static void printRanges(int[] arr, int n, int K)
{
int i = 0, j = 0, f = 0;
for (i = 0; i < n; i++) {
int mx = Int32.MinValue, mn = Int32.MaxValue;
for (j = i; j < n; j++) {
// Calculate max and min for
// any particular index range
mx = Math.Max(mx, arr[j]);
mn = Math.Min(mn, arr[j]);
// If we get max-min = K
// print 1 based index
if (mx - mn == K) {
Console.WriteLine((i + 1) + " "
+ (j + 1));
f = 1;
}
}
}
// If we didn't find any index ranges
if (f == 0)
Console.Write(-1);
}
// Driver Code
public static void Main()
{
int[] arr = { 2, 1, 3, 4, 2, 6 };
int N = arr.Length;
int K = 2;
// Function call
printRanges(arr, N, K);
}
}
// This code is contributed by Samim Hossain Mondal.
<script>
// Javascript implementation of above approach
// Function to print index ranges
function printRanges(arr, n, K)
{
let i = 0, j = 0, f = 0;
for (i = 0; i < n; i++) {
let mx = Number.MIN_SAFE_INTEGER, mn = Number.MAX_SAFE_INTEGER;
for (j = i; j < n; j++) {
// Calculate max and min for
// any particular index range
mx = Math.max(mx, arr[j]);
mn = Math.min(mn, arr[j]);
// If we get max-min = K
// print 1 based index
if (mx - mn == K) {
document.write((i + 1) + " " + (j + 1));
f = 1;
}
}
}
// If we didn't find any index ranges
if (f == 0)
document.write(-1);
}
// Driver Code
let arr = [ 2, 1, 3, 4, 2, 6 ];
let N = arr.length;
let K = 2;
// Function call
printRanges(arr, N, K);
// This code is contributed by Samim Hossain Mondal.
</script>
Time Complexity: O(N2)
Auxiliary Space: O(1)
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