Given an array of length N and an integer K. The task is to find the maximum length L such that all the subarrays of length L have sum of its elements less than K.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}, K = 20
Output: 5
The only subarray of length 5 is the complete
array and (1 + 2 + 3 + 4 + 5) = 15 < 20.Input: arr[] = {1, 2, 3, 4, 5}, K = 10
Output: 2
Approach: For the maximum sum of a subarray of length K, go through the approach discussed in this article. Now, a binary search can be performed to find the maximum length. As the array elements are positive then increasing the subarray length will increase the maximum sum of the subarray elements for that length.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;
// Function to return the maximum sum// in a subarray of size kint maxSum(int arr[], int n, int k)
{ // k must be greater
if (n < k) {
return -1;
}
// Compute sum of first window of size k
int res = 0;
for (int i = 0; i < k; i++)
res += arr[i];
// Compute sums of remaining windows by
// removing first element of previous
// window and adding last element of
// current window.
int curr_sum = res;
for (int i = k; i < n; i++) {
curr_sum += arr[i] - arr[i - k];
res = max(res, curr_sum);
}
return res;
}// Function to return the length of subarray// Sum of all the subarray of this// length is less than or equal to Kint solve(int arr[], int n, int k)
{ int max_len = 0, l = 0, r = n, m;
// Binary search from l to r as all the
// array elements are positive so that
// the maximum subarray sum is monotonically
// increasing
while (l <= r) {
m = (l + r) / 2;
// Check if the subarray sum is
// greater than K or not
if (maxSum(arr, n, m) > k)
r = m - 1;
else {
l = m + 1;
// Update the maximum length
max_len = m;
}
}
return max_len;
}// Driver codeint main()
{ int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr) / sizeof(int);
int k = 10;
cout << solve(arr, n, k);
return 0;
} |
Java
// Java implementation of the approach class GFG
{ // Function to return the maximum sum
// in a subarray of size k
static int maxSum(int arr[], int n, int k)
{
// k must be greater
if (n < k)
{
return -1;
}
// Compute sum of first window of size k
int res = 0;
for (int i = 0; i < k; i++)
res += arr[i];
// Compute sums of remaining windows by
// removing first element of previous
// window and adding last element of
// current window.
int curr_sum = res;
for (int i = k; i < n; i++)
{
curr_sum += arr[i] - arr[i - k];
res = Math.max(res, curr_sum);
}
return res;
}
// Function to return the length of subarray
// Sum of all the subarray of this
// length is less than or equal to K
static int solve(int arr[], int n, int k)
{
int max_len = 0, l = 0, r = n, m;
// Binary search from l to r as all the
// array elements are positive so that
// the maximum subarray sum is monotonically
// increasing
while (l <= r)
{
m = (l + r) / 2;
// Check if the subarray sum is
// greater than K or not
if (maxSum(arr, n, m) > k)
r = m - 1;
else
{
l = m + 1;
// Update the maximum length
max_len = m;
}
}
return max_len;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = arr.length;
int k = 10;
System.out.println(solve(arr, n, k));
}
}// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach # Function to return the maximum sum # in a subarray of size k def maxSum(arr, n, k) :
# k must be greater
if (n < k) :
return -1;
# Compute sum of first window of size k
res = 0;
for i in range(k) :
res += arr[i];
# Compute sums of remaining windows by
# removing first element of previous
# window and adding last element of
# current window.
curr_sum = res;
for i in range(k, n) :
curr_sum += arr[i] - arr[i - k];
res = max(res, curr_sum);
return res;
# Function to return the length of subarray # Sum of all the subarray of this # length is less than or equal to K def solve(arr, n, k) :
max_len = 0; l = 0; r = n;
# Binary search from l to r as all the
# array elements are positive so that
# the maximum subarray sum is monotonically
# increasing
while (l <= r) :
m = (l + r) // 2;
# Check if the subarray sum is
# greater than K or not
if (maxSum(arr, n, m) > k) :
r = m - 1;
else :
l = m + 1;
# Update the maximum length
max_len = m;
return max_len;
# Driver code if __name__ == "__main__" :
arr = [ 1, 2, 3, 4, 5 ];
n = len(arr);
k = 10;
print(solve(arr, n, k));
# This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System;
class GFG
{ // Function to return the maximum sum
// in a subarray of size k
static int maxSum(int []arr, int n, int k)
{
// k must be greater
if (n < k)
{
return -1;
}
// Compute sum of first window of size k
int res = 0;
for (int i = 0; i < k; i++)
res += arr[i];
// Compute sums of remaining windows by
// removing first element of previous
// window and adding last element of
// current window.
int curr_sum = res;
for (int i = k; i < n; i++)
{
curr_sum += arr[i] - arr[i - k];
res = Math.Max(res, curr_sum);
}
return res;
}
// Function to return the length of subarray
// Sum of all the subarray of this
// length is less than or equal to K
static int solve(int []arr, int n, int k)
{
int max_len = 0, l = 0, r = n, m;
// Binary search from l to r as all the
// array elements are positive so that
// the maximum subarray sum is monotonically
// increasing
while (l <= r)
{
m = (l + r) / 2;
// Check if the subarray sum is
// greater than K or not
if (maxSum(arr, n, m) > k)
r = m - 1;
else
{
l = m + 1;
// Update the maximum length
max_len = m;
}
}
return max_len;
}
// Driver code
public static void Main ()
{
int []arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
int k = 10;
Console.WriteLine(solve(arr, n, k));
}
}// This code is contributed by AnkitRai01 |
Javascript
<script>// javascript implementation of the approach // Function to return the maximum sum// in a subarray of size k function maxSum(arr , n , k) {
// k must be greater
if (n < k) {
return -1;
}
// Compute sum of first window of size k
var res = 0;
for (i = 0; i < k; i++)
res += arr[i];
// Compute sums of remaining windows by
// removing first element of previous
// window and adding last element of
// current window.
var curr_sum = res;
for (i = k; i < n; i++) {
curr_sum += arr[i] - arr[i - k];
res = Math.max(res, curr_sum);
}
return res;
}
// Function to return the length of subarray
// Sum of all the subarray of this
// length is less than or equal to K
function solve(arr , n , k) {
var max_len = 0, l = 0, r = n, m;
// Binary search from l to r as all the
// array elements are positive so that
// the maximum subarray sum is monotonically
// increasing
while (l <= r) {
m = parseInt((l + r) / 2);
// Check if the subarray sum is
// greater than K or not
if (maxSum(arr, n, m) > k)
r = m - 1;
else {
l = m + 1;
// Update the maximum length
max_len = m;
}
}
return max_len;
}
// Driver code
var arr = [ 1, 2, 3, 4, 5 ];
var n = arr.length;
var k = 10;
document.write(solve(arr, n, k));
// This code contributed by Rajput-Ji</script> |
Output
2
Time Complexity: O(N*logN), as we are using binary search which will cost logN and in each traversal, we are calling the function maxSum which will cost O(N) time. Where N is the number of elements in the array.
Auxiliary Space: O(1), as we are not using any extra space.