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Maximum subarray size, such that all subarrays of that size have sum less than k
  • Difficulty Level : Hard
  • Last Updated : 22 Dec, 2020

Given an array of n positive integers and a positive integer k, the task is to find the maximum subarray size such that all subarrays of that size have sum of elements less than or equals to k.

Examples : 

Input :  arr[] = {1, 2, 3, 4} and k = 8.
Output : 2
Sum of subarrays of size 1: 1, 2, 3, 4.
Sum of subarrays of size 2: 3, 5, 7.
Sum of subarrays of size 3: 6, 9.
Sum of subarrays of size 4: 10.
So, maximum subarray size such that all subarrays
of that size have the sum of elements less than 8 is 2.

Input :  arr[] = {1, 2, 10, 4} and k = 8.
Output : -1
There is an array element with value greater than k,
so subarray sum cannot be less than k.

Input :  arr[] = {1, 2, 10, 4} and K = 14
Output : 2

Naive Approach: Firstly, the required subarray size must lie between 1 to n. Now, since all the array elements are positive integers, we can say that the prefix sum of any subarray shall be strictly increasing. Thus, we can say that 

if arr[i] + arr[i + 1] + ..... + arr[j - 1] + arr[j] <= K
then arr[i] + arr[i + 1] + ..... + arr[j - 1] <= K, as
arr[j] is a positive integer.
  • Perform Binary Search over the range 1 to n and find the highest subarray size such that all the subarrays of that size have the sum of elements less than or equals to k.

Below is the implementation of the above approach:

C++




// C++ program to find maximum
// subarray size, such that all
// subarrays of that size have
// sum less than K.
#include<bits/stdc++.h>
using namespace std;
 
// Search for the maximum length of
// required subarray.
int bsearch(int prefixsum[], int n,
                             int k)
{
    // Initialize result
    int ans = -1;
 
    // Do Binary Search for largest
    // subarray size
    int left = 1, right = n;
    while (left <= right)
    {
        int mid = (left + right) / 2;
 
        // Check for all subarrays after mid
        int i;
        for (i = mid; i <= n; i++)
        {
            // Checking if all the subarrays
            //  of a size less than k.
            if (prefixsum[i] - prefixsum[i - mid] > k)
                break;
        }
 
        // All subarrays of size mid have
        // sum less than or equal to k
        if (i == n + 1)
        {
            left = mid + 1;
            ans = mid;
        }
 
        // We found a subrray of size mid
        // with sum greater than k
        else
            right = mid - 1;
    }
    return ans;
}
 
// Return the maximum subarray size,
// such that all subarray of that size
// have sum less than K.
int maxSize(int arr[], int n, int k)
{
    // Initialize prefix sum array as 0.
    int prefixsum[n + 1];
    memset(prefixsum, 0, sizeof(prefixsum));
 
    // Finding prefix sum of the array.
    for (int i = 0; i < n; i++)
        prefixsum[i + 1] = prefixsum[i] +
                           arr[i];
 
    return bsearch(prefixsum, n, k);
}
 
// Driver code
int main()
{
    int arr[] = {1, 2, 10, 4};
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 14;
    cout << maxSize(arr, n, k) << endl;
    return 0;
}


Java




// Java program to find maximum
// subarray size, such that all
// subarrays of that size have
// sum less than K.
import java.util.Arrays;
 
class GFG
{
     
    // Search for the maximum length
    // of required subarray.
    static int bsearch(int prefixsum[],
                       int n, int k)
    {
        // Initialize result
        int ans = -1;
 
        // Do Binary Search for largest
        // subarray size
        int left = 1, right = n;
         
        while (left <= right)
        {
            int mid = (left + right) / 2;
 
            // Check for all subarrays after mid
            int i;
            for (i = mid; i <= n; i++)
            {
                 
                // Checking if all the subarrays
                // of a size is less than k.
                if (prefixsum[i] - prefixsum[i - mid] > k)
                    break;
            }
 
            // All subarrays of size mid have
            // sum less than or equal to k
            if (i == n + 1)
            {
                left = mid + 1;
                ans = mid;
            }
 
            // We found a subrray of size mid
            // with sum greater than k
            else
                right = mid - 1;
        }
 
        return ans;
    }
 
    // Return the maximum subarray size, such
    // that all subarray of that size have
    // sum less than K.
    static int maxSize(int arr[], int n, int k)
    {
         
        // Initialize prefix sum array as 0.
        int prefixsum[] = new int[n + 1];
        Arrays.fill(prefixsum, 0);
 
        // Finding prefix sum of the array.
        for (int i = 0; i < n; i++)
            prefixsum[i + 1] = prefixsum[i] + arr[i];
 
        return bsearch(prefixsum, n, k);
    }
     
    // Driver code
    public static void main(String arg[])
    {
        int arr[] = { 1, 2, 10, 4 };
        int n = arr.length;
        int k = 14;
        System.out.println(maxSize(arr, n, k));
    }
}
 
// This code is contributed by Anant Agarwal.


Python3




# Python3 program to find maximum
# subarray size, such that all
# subarrays of that size have
# sum less than K.
 
# Search for the maximum length of
# required subarray.
def bsearch(prefixsum, n, k):
 
    # Initialize result
    # Do Binary Search for largest
    # subarray size
    ans, left, right = -1, 1, n
 
    while (left <= right):
 
        # Check for all subarrays after mid
        mid = (left + right)//2
 
        for i in range(mid, n + 1):
 
            # Checking if all the subarray of
            # a size is less than k.
            if (prefixsum[i] - prefixsum[i - mid] > k):
                i = i - 1
                break
        i = i + 1
        if (i == n + 1):
            left = mid + 1
            ans = mid
        # We found a subrray of size mid with sum
        # greater than k
        else:
            right = mid - 1
 
    return ans;
 
# Return the maximum subarray size, such
# that all subarray of that size have
# sum less than K.
def maxSize(arr, n, k):
    prefixsum = [0 for x in range(n + 1)]
     
    # Finding prefix sum of the array.
    for i in range(n):
        prefixsum[i + 1] = prefixsum[i] + arr[i]
 
    return bsearch(prefixsum, n, k);
 
# Driver Code
arr = [ 1, 2, 10, 4 ]
n = len(arr)
k = 14
print (maxSize(arr, n, k))
 
# This code is contributed by Afzal


C#




// C# program to find maximum
// subarray size, such that all
// subarrays of that size have
// sum less than K.
using System;
 
class GFG {
     
    // Search for the maximum length
    // of required subarray.
    static int bsearch(int []prefixsum,
                          int n, int k)
    {
         
        // Initialize result
        int ans = -1;
 
        // Do Binary Search for
        // largest subarray size
        int left = 1, right = n;
         
        while (left <= right)
        {
            int mid = (left + right) / 2;
 
            // Check for all subarrays
            // after mid
            int i;
            for (i = mid; i <= n; i++)
            {
                 
                // Checking if all the
                // subarrays of a size is
                // less than k.
                if (prefixsum[i] -
                     prefixsum[i - mid] > k)
                    break;
            }
 
            // All subarrays of size mid have
            // sum less than or equal to k
            if (i == n + 1)
            {
                left = mid + 1;
                ans = mid;
            }
 
            // We found a subrray of size mid
            // with sum greater than k
            else
                right = mid - 1;
        }
 
        return ans;
    }
 
    // Return the maximum subarray size, such
    // that all subarray of that size have
    // sum less than K.
    static int maxSize(int []arr, int n, int k)
    {
         
        // Initialize prefix sum array as 0.
        int []prefixsum = new int[n + 1];
        for(int i=0;i<n+1;i++)
        prefixsum[i]=0;
         
 
        // Finding prefix sum of the array.
        for (int i = 0; i < n; i++)
            prefixsum[i + 1] = prefixsum[i]
                                     + arr[i];
 
        return bsearch(prefixsum, n, k);
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = { 1, 2, 10, 4 };
        int n = arr.Length;
        int k = 14;
         
        Console.Write(maxSize(arr, n, k));
    }
}
 
// This code is contributed by nitin mittal.


PHP




<?php
// PHP program to find maximum subarray
// size, such that all subarrays of that
// size have sum less than K.
 
// Search for the maximum length of
// required subarray.
function bsearch(&$prefixsum, $n, $k)
{
    // Initialize result
    $ans = -1;
 
    // Do Binary Search for largest
    // subarray size
    $left = 1;
    $right = $n;
    while ($left <= $right)
    {
        $mid = intval(($left + $right) / 2);
 
        // Check for all subarrays after mid
        for ($i = $mid; $i <= $n; $i++)
        {
            // Checking if all the subarrays
            // of a size less than k.
            if ($prefixsum[$i] - $prefixsum[$i -
                                 $mid] > $k)
                break;
        }
 
        // All subarrays of size mid have
        // sum less than or equal to k
        if ($i == $n + 1)
        {
            $left = $mid + 1;
            $ans = $mid;
        }
 
        // We found a subrray of size mid
        // with sum greater than k
        else
            $right = $mid - 1;
    }
    return $ans;
}
 
// Return the maximum subarray size,
// such that all subarray of that size
// have sum less than K.
function maxSize(&$arr, $n, $k)
{
    // Initialize prefix sum array as 0.
    $prefixsum = array_fill(0, $n + 1, NULL);
 
    // Finding prefix sum of the array.
    for ($i = 0; $i < $n; $i++)
        $prefixsum[$i + 1] = $prefixsum[$i] +
                             $arr[$i];
 
    return bsearch($prefixsum, $n, $k);
}
 
// Driver code
$arr = array(1, 2, 10, 4);
$n = sizeof($arr);
$k = 14;
echo maxSize($arr, $n, $k) . "\n";
 
// This code is contributed
// by ChitraNayal
?>


Output



2

Time Complexity: O(n log n)

Efficient Approach: This method uses the Sliding Window Technique to solve the given problem. 

  • The approach is to find the minimum subarray size whose sum is greater than integer k.
  • Increase the window size from the end up to which the sum of that window is greater than k.
  • Now, store that subarray size if it is smaller than already stored subarray size (in variable ans).
  • Now, decrement the subarray size from the beginning. The variable ans will store the minimum subarray size whose sum is greater than k.
  • At last, (ans-1) is the actual answer. Then, that subarray size – 1 is the maximum subarray size, such that all subarray of that size will have sum less than or equal to k.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// largest size subarray
void func(vector<int> arr,
          int k, int n)
{
    // Variable declaration
    int ans = n;
    int sum = 0;
    int start = 0;
 
    // Loop till N
    for (int end = 0; end < n; end++)
    {
        // Sliding window from left
        sum += arr[end];
 
        while (sum > k) {
            // Sliding window from right
            sum -= arr[start];
            start++;
 
            // Storing sub-array size - 1
            // for which sum was greater than k
            ans = min(ans, end - start + 1);
 
            // Sum will be 0 if start>end
            // because all elements are positive
            // start>end only when arr[end]>k i.e,
            // there is an array element with
            // value greater than k, so sub-array
            // sum cannot be less than k.
            if (sum == 0)
                break;
        }
        if (sum == 0) {
            ans = -1;
            break;
        }
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver code
int main()
{
    vector<int> arr{ 1, 2, 3, 4 };
    int k = 8;
    int n = arr.size();
 
    // Function call
    func(arr, k, n);
 
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to find the
// largest size subarray
public static void func(int arr[],
                        int k, int n)
{
     
    // Variable declaration
    int ans = n;
    int sum = 0;
    int start = 0;
     
    // Loop till N
    for(int end = 0; end < n; end++)
    {
         
        // Sliding window from left
        sum += (int)arr[end];
         
        while (sum > k)
        {
             
            // Sliding window from right
            sum -= (int)arr[start];
            start++;
 
            // Storing sub-array size - 1
            // for which sum was greater than k
            ans = Math.min(ans, end - start + 1);
 
            // Sum will be 0 if start>end
            // because all elements are positive
            // start>end only when arr[end]>k i.e,
            // there is an array element with
            // value greater than k, so sub-array
            // sum cannot be less than k.
            if (sum == 0)
                break;
        }
         
        if (sum == 0)
        {
            ans = -1;
            break;
        }
    }
     
    // Print the answer
    System.out.println(ans);
}
 
// Driver code
public static void main (String[] args)
{
    int arr[] = { 1, 2, 3, 4 };
    int k = 8;
    int n = arr.length;
     
    // Function call
    func(arr, k, n);
}
}
 
// This code is contributed by rag2127


Python3




# Python3 program for the above approach
 
# Function to find the
# largest size subarray
def func(arr, k, n):
     
    # Variable declaration
    ans = n
    Sum = 0
    start = 0
 
    # Loop till N
    for end in range(n):
 
        # Sliding window from left
        Sum += arr[end]
 
        while (Sum > k):
             
            # Sliding window from right
            Sum -= arr[start]
            start += 1
 
            # Storing sub-array size - 1
            # for which sum was greater than k
            ans = min(ans, end - start + 1)
 
            # Sum will be 0 if start>end
            # because all elements are positive
            # start>end only when arr[end]>k i.e,
            # there is an array element with
            # value greater than k, so sub-array
            # sum cannot be less than k.
            if (Sum == 0):
                break
                 
        if (Sum == 0):
            ans = -1
            break
             
    # Print the answer
    print(ans)
 
# Driver code
arr = [ 1, 2, 3, 4 ]
k = 8
n = len(arr)
 
# Function call
func(arr, k, n)
 
# This code is contributed by avanitrachhadiya2155


C#




// C# program for the above approach
using System;
using System.Collections;
 
class GFG{
      
// Function to find the
// largest size subarray
static void func(ArrayList arr,
                 int k, int n)
{
     
    // Variable declaration
    int ans = n;
    int sum = 0;
    int start = 0;
  
    // Loop till N
    for(int end = 0; end < n; end++)
    {
         
        // Sliding window from left
        sum += (int)arr[end];
  
        while (sum > k)
        {
             
            // Sliding window from right
            sum -= (int)arr[start];
            start++;
  
            // Storing sub-array size - 1
            // for which sum was greater than k
            ans = Math.Min(ans, end - start + 1);
  
            // Sum will be 0 if start>end
            // because all elements are positive
            // start>end only when arr[end]>k i.e,
            // there is an array element with
            // value greater than k, so sub-array
            // sum cannot be less than k.
            if (sum == 0)
                break;
        }
        if (sum == 0)
        {
            ans = -1;
            break;
        }
    }
     
    // Print the answer
    Console.Write(ans);
}
  
// Driver code
public static void Main(string[] args)
{
    ArrayList arr = new ArrayList(){ 1, 2, 3, 4 };
    int k = 8;
    int n = arr.Count;
  
    // Function call
    func(arr, k, n);
}
}
 
// This code is contributed by rutvik_56


Output

2

Time Complexity: O(N)

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