Minimum Operations to Reduce X to Zero - Practice Coding Problems

Minimum Operations to Reduce X to Zero - Problem

Imagine you have an array of integers and a target number x. Your goal is to reduce x to exactly 0 using the minimum number of operations.

In each operation, you can choose to remove either the leftmost or rightmost element from the array and subtract its value from x. Once an element is removed, the array shrinks and future operations can only work with the remaining elements.

The Challenge: Find the minimum number of operations needed to make x exactly zero. If it's impossible, return -1.

Example: If nums = [1,1,4,2,3] and x = 5, you could remove 1 (left) and 4 (next left) to get 1 + 4 = 5, requiring 2 operations.

Input & Output

example_1.py — Basic Case

$ Input: nums = [1,1,4,2,3], x = 5

› Output: 2

💡 Note: Remove the first element (1) and the last element (3), giving us 1 + 4 = 5. This takes 2 operations and is optimal.

example_2.py — Remove All Elements

$ Input: nums = [5,6,7,8,9], x = 35

› Output: 5

💡 Note: The sum of all elements is 35, which equals x. We need to remove all elements, requiring 5 operations.

example_3.py — Impossible Case

$ Input: nums = [3,2,20,1,1,3], x = 10

› Output: 5

💡 Note: We can remove [3,2] from left and [1,1,3] from right to get 3+2+1+1+3=10. Total operations: 5.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁴
1 ≤ x ≤ 2 * 10⁹
All elements are positive integers

Visualization

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Understanding the Visualization

Calculate total

Sum all cookie values: 1+1+4+2+3 = 11

Find target

Middle section should sum to: 11 - 5 = 6

Use sliding window

Find longest contiguous sequence summing to 6

Calculate answer

Operations needed = total_length - longest_middle_sequence

Key Takeaway

🎯 Key Insight: Transform the problem from "what to remove from ends" to "what longest sequence to keep in middle" - this changes O(n²) brute force to elegant O(n) sliding window!

Asked in

f Facebook 45 a Amazon 38 G Google 32 ⊞ Microsoft 28

The optimal solution uses a sliding window approach with a key insight: instead of thinking about what to remove from the ends, find the longest contiguous subarray to keep in the middle. If the total sum is S and we need to remove elements summing to x, then we need a subarray with sum = S - x. The answer is n - (length of longest valid subarray). This transforms an O(n²) problem into an elegant O(n) solution.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(n²)	O(1)	Try all possible combinations of removing elements from left and right ends
Sliding Window (Optimal)	O(n)	O(1)	Transform problem to finding maximum subarray with sum = total - x

Brute Force (Try All Combinations) — Algorithm Steps

For each count i from 0 to n, take i elements from the left
Calculate the sum of these i left elements
Calculate remaining needed: x - leftSum
Try to achieve remaining by taking elements from right end
Track minimum operations across all valid combinations

Visualization

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Step-by-Step Walkthrough

Try 0 from left

Check if we can get sum x using only right elements

Try 1 from left

Take first element, check remaining sum from right

Try 2 from left

Take first two elements, check remaining from right

Continue pattern

Keep trying until all combinations are explored

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int minimumOperations(int* nums, int numsSize, int x) {
    int minOps = INT_MAX;
    
    // Try all possible combinations of left and right elements
    for (int leftCount = 0; leftCount <= numsSize; leftCount++) {
        int leftSum = 0;
        for (int i = 0; i < leftCount; i++) {
            leftSum += nums[i];
        }
        
        if (leftSum == x) {
            if (leftCount < minOps) {
                minOps = leftCount;
            }
            continue;
        }
        
        if (leftSum > x) break;
        
        int remaining = x - leftSum;
        int rightSum = 0;
        
        for (int rightCount = 1; rightCount <= numsSize - leftCount; rightCount++) {
            rightSum += nums[numsSize - rightCount];
            
            if (rightSum == remaining) {
                int total = leftCount + rightCount;
                if (total < minOps) {
                    minOps = total;
                }
                break;
            } else if (rightSum > remaining) {
                break;
            }
        }
    }
    
    return minOps == INT_MAX ? -1 : minOps;
}

int main() {
    int nums[] = {1, 1, 4, 2, 3};
    int x = 5;
    printf("%d\n", minimumOperations(nums, 5, x));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n positions from left, we potentially scan up to n positions from right

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track sums and minimum operations

✓ Linear Space

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High Frequency

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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