# Minimum number of elements to be removed such that the sum of the remaining elements is equal to k

Given an array arr[] of integers and an integer k, the task is to find the minimum number of integers that need to be removed from the array such that the sum of the remaining elements is equal to k. If we cannot get the required sum the print -1.

Examples:

Input: arr[] = {1, 2, 3}, k = 3
Output: 1
Either remove 1 and 2 to reduce the array to {3}
or remove 3 to get the array {1, 2}. Both have equal sum i.e. 3
But removing 3 requires only a single removal.

Input: arr[] = {1, 3, 2, 5, 6}, k = 5
Output: 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to use a sliding window and variables j initialize it to 0, min_num to store the answer and sum to store the current sum. Keep on adding the elements of the array to the variable sum, till it becomes greater than or equal to k, if it is equal to k, then update the min_num as minimum of min_num and (n -(i+1) +j) where n is the number of integers in array and i is the current index, else if it is greater than k, then start decrementing the sum by removing the values of the array from sum till the sum becomes less than or equal to k and also increment the value of j, if sum is equal to k, then once again update min_num. Repeat this whole process till the end of the array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum number of ` `// integers that need to be removed from the ` `// array to form a sub-array with sum k ` `int` `FindMinNumber(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``int` `i = 0; ` `    ``int` `j = 0; ` ` `  `    ``// Stores the minimum number of ` `    ``// integers that need to be removed ` `    ``// from the array ` `    ``int` `min_num = INT_MAX; ` ` `  `    ``bool` `found = ``false``; ` ` `  `    ``int` `sum = 0; ` ` `  `    ``while` `(i < n) { ` ` `  `        ``sum = sum + arr[i]; ` ` `  `        ``// If current sum is equal to ` `        ``// k, update min_num ` `        ``if` `(sum == k) { ` `            ``min_num = min(min_num, ((n - (i + 1)) + j)); ` `            ``found = ``true``; ` `        ``} ` ` `  `        ``// If current sum is greater than k ` `        ``else` `if` `(sum > k) { ` ` `  `            ``// Decrement the sum until it ` `            ``// becomes less than or equal to k ` `            ``while` `(sum > k) { ` `                ``sum = sum - arr[j]; ` `                ``j++; ` `            ``} ` `            ``if` `(sum == k) { ` `                ``min_num = min(min_num, ((n - (i + 1)) + j)); ` `                ``found = ``true``; ` `            ``} ` `        ``} ` ` `  `        ``i++; ` `    ``} ` ` `  `    ``if` `(found) ` `        ``return` `min_num; ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 3, 2, 5, 6 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` `    ``int` `k = 5; ` ` `  `    ``cout << FindMinNumber(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `     `  `// Function to return the minimum number of ` `// integers that need to be removed from the ` `// array to form a sub-array with sum k ` `static` `int` `FindMinNumber(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``int` `i = ``0``; ` `    ``int` `j = ``0``; ` ` `  `    ``// Stores the minimum number of ` `    ``// integers that need to be removed ` `    ``// from the array ` `    ``int` `min_num = Integer.MAX_VALUE; ` ` `  `    ``boolean` `found = ``false``; ` ` `  `    ``int` `sum = ``0``; ` ` `  `    ``while` `(i < n)  ` `    ``{ ` ` `  `        ``sum = sum + arr[i]; ` ` `  `        ``// If current sum is equal to ` `        ``// k, update min_num ` `        ``if` `(sum == k) ` `        ``{ ` `            ``min_num = Math.min(min_num,  ` `                             ``((n - (i + ``1``)) + j)); ` `            ``found = ``true``; ` `        ``} ` ` `  `        ``// If current sum is greater than k ` `        ``else` `if` `(sum > k)  ` `        ``{ ` ` `  `            ``// Decrement the sum until it ` `            ``// becomes less than or equal to k ` `            ``while` `(sum > k)  ` `            ``{ ` `                ``sum = sum - arr[j]; ` `                ``j++; ` `            ``} ` `            ``if` `(sum == k)  ` `            ``{ ` `                ``min_num = Math.min(min_num,  ` `                                 ``((n - (i + ``1``)) + j)); ` `                ``found = ``true``; ` `            ``} ` `        ``} ` ` `  `        ``i++; ` `    ``} ` ` `  `    ``if` `(found) ` `        ``return` `min_num; ` ` `  `    ``return` `-``1``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``3``, ``2``, ``5``, ``6` `}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``5``; ` ` `  `    ``System.out.println(FindMinNumber(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the minimum number of ` `# integers that need to be removed from the ` `# array to form a sub-array with Sum k ` `def` `FindMinNumber(arr, n, k): ` `    ``i ``=` `0` `    ``j ``=` `0` ` `  `    ``# Stores the minimum number of ` `    ``# integers that need to be removed ` `    ``# from the array ` `    ``min_num ``=` `10``*``*``9` ` `  `    ``found ``=` `False` ` `  `    ``Sum` `=` `0` ` `  `    ``while` `(i < n): ` ` `  `        ``Sum` `=` `Sum` `+` `arr[i] ` ` `  `        ``# If current Sum is equal to ` `        ``# k, update min_num ` `        ``if` `(``Sum` `=``=` `k): ` `            ``min_num ``=` `min``(min_num, ` `                        ``((n ``-` `(i ``+` `1``)) ``+` `j)) ` `            ``found ``=` `True` `         `  `        ``# If current Sum is greater than k ` `        ``elif` `(``Sum` `> k): ` ` `  `            ``# Decrement the Sum until it ` `            ``# becomes less than or equal to k ` `            ``while` `(``Sum` `> k): ` `                ``Sum` `=` `Sum` `-` `arr[j] ` `                ``j ``+``=` `1` `            ``if` `(``Sum` `=``=` `k): ` `                ``min_num ``=` `min``(min_num,  ` `                            ``((n ``-` `(i ``+` `1``)) ``+` `j)) ` `                ``found ``=` `True` `             `  `        ``i ``+``=` `1` ` `  `    ``if` `(found): ` `        ``return` `min_num ` ` `  `    ``return` `-``1` ` `  `# Driver code ` `arr ``=` `[``1``, ``3``, ``2``, ``5``, ``6``] ` `n ``=` `len``(arr) ` `k ``=` `5` ` `  `print``(FindMinNumber(arr, n, k)) ` ` `  `# This code is contributed by mohit kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the minimum number of ` `// integers that need to be removed from the ` `// array to form a sub-array with sum k ` `static` `int` `FindMinNumber(``int``[] arr, ``int` `n, ``int` `k) ` `{ ` `    ``int` `i = 0; ` `    ``int` `j = 0; ` ` `  `    ``// Stores the minimum number of ` `    ``// integers that need to be removed ` `    ``// from the array ` `    ``int` `min_num = ``int``.MaxValue; ` ` `  `    ``bool` `found = ``false``; ` ` `  `    ``int` `sum = 0; ` ` `  `    ``while` `(i < n)  ` `    ``{ ` ` `  `        ``sum = sum + arr[i]; ` ` `  `        ``// If current sum is equal to ` `        ``// k, update min_num ` `        ``if` `(sum == k) ` `        ``{ ` `            ``min_num = Math.Min(min_num,  ` `                            ``((n - (i + 1)) + j)); ` `            ``found = ``true``; ` `        ``} ` ` `  `        ``// If current sum is greater than k ` `        ``else` `if` `(sum > k)  ` `        ``{ ` ` `  `            ``// Decrement the sum until it ` `            ``// becomes less than or equal to k ` `            ``while` `(sum > k)  ` `            ``{ ` `                ``sum = sum - arr[j]; ` `                ``j++; ` `            ``} ` `            ``if` `(sum == k)  ` `            ``{ ` `                ``min_num = Math.Min(min_num,  ` `                                ``((n - (i + 1)) + j)); ` `                ``found = ``true``; ` `            ``} ` `        ``} ` ` `  `        ``i++; ` `    ``} ` ` `  `    ``if` `(found) ` `        ``return` `min_num; ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int``[] arr = { 1, 3, 2, 5, 6 }; ` `    ``int` `n = arr.Length; ` `    ``int` `k = 5; ` ` `  `    ``Console.WriteLine(FindMinNumber(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

## PHP

 ` ``\$k``)  ` `        ``{ ` ` `  `            ``// Decrement the sum until it ` `            ``// becomes less than or equal to k ` `            ``while` `(``\$sum` `> ``\$k``)  ` `            ``{ ` `                ``\$sum` `= ``\$sum` `- ``\$arr``[``\$j``]; ` `                ``\$j``++; ` `            ``} ` `            ``if` `(``\$sum` `== ``\$k``)  ` `            ``{ ` `                ``\$min_num` `=min(``\$min_num``,  ` `                                ``((``\$n` `- (``\$i` `+ 1)) + ``\$j``)); ` `                ``\$found` `= true; ` `            ``} ` `        ``} ` ` `  `        ``\$i``++; ` `    ``} ` ` `  `    ``if` `(``\$found``) ` `        ``return` `\$min_num``; ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver code ` `\$arr` `= ``array``( 1, 3, 2, 5, 6 ); ` `\$n` `= sizeof(``\$arr``); ` `\$k` `= 5; ` ` `  `echo``(FindMinNumber(``\$arr``, ``\$n``, ``\$k``)); ` ` `  `// This code is contributed by Code_Mech `

Output:

```3
```

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Improved By : mohit kumar 29, Code_Mech