# Maximum subarray size, such that all subarrays of that size have sum less than k

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++ program to find maximum ` `// subarray size, such that all ` `// subarrays of that size have ` `// sum less than K.` `#include` `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 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.`

 `# Python 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# 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

 ` ``\$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++ program for the above approach` `#include ` `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;` `}`

 `// 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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