Minimum number of array elements from either ends required to be subtracted from X to reduce X to 0

Namachila November 18, 2025

0 0 7 minutes read

Given an array nums[] and an integer X, the task is to reduce X to 0 by removing either the leftmost or the rightmost array elements and subtracting its value from X, minimum number of times. If it’s possible to reduce X to 0, print the count of operations required. Otherwise, return -1.

Examples:

Input: nums[] = {3,2,20,1,1,3}, X = 10
Output: 5
Explanation: X (= 10) – 3 – 1 – 1 – 3 – 2 = 0. Therefore, the number of operations required is 5.

Input: nums[] = {1, 1, 4, 2, 3}, X = 5
Output: 2
Explanation: X (= 5) – 3 – 2 = 0. Therefore, the number of operations required is 2.

Approach: The given problem can be solved using Two Pointers technique. Follow the steps below to solve the problem.

Maintain two pointers left and right, pointing to the ends of the left and right subarrays excluded from X.
Initialize left to consider the entire array, and right to include nothing.
Therefore, reduce X by the sum of the array.
Now, iterate until left reaches the front of the array.
- If the sum of the left and the right subarrays exceeds X (i.e. X < 0), shift left by an index to the left and increase X that element.
- If the sum of the left and the right subarrays is less than X (i.e. X > 0), shift right pointer by an index to the left and reduce X by that element.
- If X is found to be 0 at any point, update the minimum number of operations required.
Print the minimum number of operations required.
Below is the implementation of the above approach:

C++14

// C++ program to implement 
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the minimum
// number of operations required
// to reduce x to 0
static int minOperations(int nums[], int N,
                         int x)
{
     
    // If sum of the array
    // is less than x
    int sum = 0;
     
    for(int i = 0; i < x; i++)
        sum += nums[i];
         
    if (sum < x)
        return -1;
     
    // Stores the count
    // of operations
    int ans = INT_MAX;
     
    // Two pointers to traverse the array
    int l = N - 1, r = N;
     
    // Reduce x by the sum
    // of the entire array
    x -= sum;
     
    // Iterate until l reaches
    // the front of the array
    while (l >= 0)
    {
     
        // If sum of elements from
        // the front exceeds x
        if (x <= 0) 
        {
         
            // Shift towards left
            x += nums[l];
            l -= 1;
        }
         
        // If sum exceeds 0
        if (x > 0)
        {
         
            // Reduce x by elements
            // from the right
            r -= 1;
            x -= nums[r];
        }
         
        // If x is reduced to 0
        if (x == 0)
        {
         
            // Update the minimum count
            // of operations required
            ans = min(ans,
            (l + 1) + (N - r));
        }
    }
     
    if (ans < INT_MAX)
        return ans;
    else
        return -1;
}
 
// Driver Code
int main()
{
    int nums[] = { 1, 1, 4, 2, 3 };
     
     // Size of the array
    int N = sizeof(nums) / sizeof(nums[0]);
     
    int x = 5;
    cout << minOperations(nums, N, x);
 
    return 0;
}
 
// This code is contributed by code_hunt

Java

// Java program to implement 
// the above approach 
import java.util.*;
 
class GFG
{
 
  // Function to count the minimum
  // number of operations required
  // to reduce x to 0
  static int minOperations(int nums[], int x)
  {
 
    // If sum of the array
    // is less than x
    int sum = 0;
    for (int i = 0; i < x; i++)
      sum += nums[i];
    if (sum < x)
      return -1;
 
    // Stores the count
    // of operations
    int ans = Integer.MAX_VALUE;
 
    // Two pointers to traverse the array
    int l = nums.length - 1, r = nums.length;
 
    // Reduce x by the sum
    // of the entire array
    x -= sum;
 
    // Iterate until l reaches
    // the front of the array
    while (l >= 0) {
 
      // If sum of elements from
      // the front exceeds x
      if (x <= 0) {
 
        // Shift towards left
        x += nums[l];
        l -= 1;
      }
 
      // If sum exceeds 0
      if (x > 0) {
 
        // Reduce x by elements
        // from the right
        r -= 1;
        x -= nums[r];
      }
 
      // If x is reduced to 0
      if (x == 0) {
 
        // Update the minimum count
        // of operations required
        ans = Math.min(ans,
                       (l + 1) + (nums.length - r));
      }
    }
    if (ans < Integer.MAX_VALUE)
      return ans;
    else
      return -1;
  }
 
  // Driver Code
  public static void main(String[] args) 
  {
    int[] nums = { 1, 1, 4, 2, 3 };
    int x = 5;
    System.out.println(minOperations(nums, x));
  }
}
 
// This code is contributed by shubhamsingh10

Python3

# Python3 Program to implement
# the above approach
 
import math
 
# Function to count the minimum
# number of operations required
# to reduce x to 0
def minOperations(nums, x):
 
    # If sum of the array
    # is less than x
    if sum(nums) < x:
        return -1
 
    # Stores the count
    # of operations
    ans = math.inf
 
    # Two pointers to traverse the array
    l, r = len(nums)-1, len(nums)
 
    # Reduce x by the sum
    # of the entire array
    x -= sum(nums)
 
    # Iterate until l reaches
    # the front of the array
    while l >= 0:
 
        # If sum of elements from
        # the front exceeds x
        if x <= 0:
 
            # Shift towards left
            x += nums[l]
            l -= 1
 
        # If sum exceeds 0
        if x > 0:
 
            # Reduce x by elements
            # from the right
            r -= 1
            x -= nums[r]
 
        # If x is reduced to 0
        if x == 0:
 
            # Update the minimum count
            # of operations required
            ans = min(ans, (l+1) + (len(nums)-r))
 
    return ans if ans < math.inf else -1
 
 
# Driver Code
nums = [1, 1, 4, 2, 3]
x = 5
print(minOperations(nums, x))

C#

// C# Program to implement
// the above approach
using System;
class GFG {
 
  // Function to count the minimum
  // number of operations required
  // to reduce x to 0
  static int minOperations(int[] nums, int x)
  {
 
    // If sum of the array
    // is less than x
    int sum = 0;
    for (int i = 0; i < x; i++)
      sum += nums[i];
    if (sum < x)
      return -1;
 
    // Stores the count
    // of operations
    int ans = Int32.MaxValue;
 
    // Two pointers to traverse the array
    int l = nums.Length - 1, r = nums.Length;
 
    // Reduce x by the sum
    // of the entire array
    x -= sum;
 
    // Iterate until l reaches
    // the front of the array
    while (l >= 0) {
 
      // If sum of elements from
      // the front exceeds x
      if (x <= 0) {
 
        // Shift towards left
        x += nums[l];
        l -= 1;
      }
 
      // If sum exceeds 0
      if (x > 0) {
 
        // Reduce x by elements
        // from the right
        r -= 1;
        x -= nums[r];
      }
 
      // If x is reduced to 0
      if (x == 0) {
 
        // Update the minimum count
        // of operations required
        ans = Math.Min(ans,
                       (l + 1) + (nums.Length - r));
      }
    }
    if (ans < Int32.MaxValue)
      return ans;
    else
      return -1;
  }
 
  // Driver Code
  public static void Main()
  {
    int[] nums = { 1, 1, 4, 2, 3 };
    int x = 5;
    Console.Write(minOperations(nums, x));
  }
}
 
// This code is contributed by ukasp.

Javascript

<script>
 
// Javascript program to implement 
// the above approach
 
// Function to count the minimum
// number of operations required
// to reduce x to 0
function minOperations(nums, x)
{
     
    // If sum of the array
    // is less than x
    let sum = 0;
    for(let i = 0; i < x; i++)
        sum += nums[i];
         
    if (sum < x)
        return -1;
     
    // Stores the count
    // of operations
    let ans = Number.MAX_VALUE;
     
    // Two pointers to traverse the array
    let l = nums.length - 1, r = nums.length;
     
    // Reduce x by the sum
    // of the entire array
    x -= sum;
     
    // Iterate until l reaches
    // the front of the array
    while (l >= 0) 
    {
     
        // If sum of elements from
        // the front exceeds x
        if (x <= 0) 
        {
             
            // Shift towards left
            x += nums[l];
            l -= 1;
        }
         
        // If sum exceeds 0
        if (x > 0) 
        {
         
            // Reduce x by elements
            // from the right
            r -= 1;
            x -= nums[r];
        }
         
        // If x is reduced to 0
        if (x == 0) 
        {
         
            // Update the minimum count
            // of operations required
            ans = Math.min(ans,
                          (l + 1) + 
                          (nums.length - r));
        }
    }
     
    if (ans < Number.MAX_VALUE)
        return ans;
    else
        return -1;
}
 
// Driver code
let nums = [ 1, 1, 4, 2, 3 ];
let x = 5;
 
document.write(minOperations(nums, x));
 
// This code is contributed by target_2
     
</script>

PHP

// PHP program to implement
<?php
 
// Function to count the minimum number
function minOperations($nums, $N, $x) {
 
    // sum of the array
    $sum = 0;
 
    for($i = 0; $i < $x; $i++)
        $sum += $nums[$i];
 
    if ($sum < $x)
        return -1;
 
    // Stores the count of operations
    $ans = PHP_INT_MAX;
 
    // Two pointers to traverse the array
    $l = $N - 1;
    $r = $N;
 
    // Reduce x by the sum
    $x -= $sum;
 
    // Iterate until l reaches
    while ($l >= 0)
    {
 
        // sum of elements from
        if ($x <= 0)
        {
 
            // Shift towards left
            $x += $nums[$l];
            $l -= 1;
        }
 
        // If sum exceeds 0
        if ($x > 0)
        {
            $r -= 1;
            $x -= $nums[$r];
        }
 
        // x is reduced to 0
        if ($x == 0)
        {
 
            // updating the minimum count
            $ans = min($ans, ($l + 1) + ($N - $r));
        }
    }
 
    if ($ans < PHP_INT_MAX)
        return $ans;
    else
        return -1;
}
 
// Driver Code
    $nums = array(1, 1, 4, 2, 3);
    $N = count($nums);
    $x = 5;
    echo minOperations($nums, $N, $x);
?>

Output:

Time Complexity: O(N)
Auxiliary Space: O(1)

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