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

05 Aug, 2021
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  
 
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++
Java
Python3
C#
PHP
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>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 kint 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 codeint 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 implementation of the approachclass GFG{     // Function to return the minimum number of// integers that need to be removed from the// array to form a sub-array with sum kstatic 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 codepublic 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 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 kdef 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 codearr = [1, 3, 2, 5, 6]n = len(arr)k = 5 print(FindMinNumber(arr, n, k)) # This code is contributed by mohit kumar
// C# implementation of the approachusing 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 kstatic 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 codepublic 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// PHP 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 kfunction FindMinNumber($arr,$n,$k){    $i = 0;    $j = 0;     // Stores the minimum number of    // integers that need to be removed    // from the array    $min_num = PHP_INT_MAX;     $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));            $found = true;        }         // If current sum is greater than k        else if ($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 $min_num;     return -1;} // Driver code$arr = array( 1, 3, 2, 5, 6 );$n = sizeof($arr);$k = 5; echo(FindMinNumber($arr, $n, $k)); // This code is contributed by Code_Mech
<script>// Javascript 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 kfunction FindMinNumber(arr,n,k){    let i = 0;    let j = 0;       // Stores the minimum number of    // integers that need to be removed    // from the array    let min_num = Number.MAX_VALUE;     let found = false;    let 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 codelet arr = [1, 3, 2, 5, 6 ];let n = arr.length;let k = 5;document.write(FindMinNumber(arr, n, k)); // This code is contributed by patel2127</script>
3
 
Time Complexity: O(N)Auxiliary Space: O(1) 
mohit kumar 29
Code_Mech
patel2127
pankajsharmagfg
sliding-window
Algorithms
Arrays
Data Structures
Data Structures
sliding-window
Arrays
Algorithms
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