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

  • Difficulty Level : Hard
  • Last Updated : 17 Nov, 2021

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:

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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>
Output: 
2

 

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




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