2749. Minimum Operations To Make The Integer Zero

LeetCode 2749 – Minimum Operations to Make the Integer Zero

Introduction
When you encounter a problem that asks for the minimum number of operations needed to turn an array of integers into all zeros, the intuitive thought is to simulate each move. On the flip side, brute‑force simulation quickly becomes infeasible for large inputs. LeetCode 2749, Minimum Operations to Make the Integer Zero, teaches us a powerful observation: the answer can be derived directly from the prefix sums of the original array. This article walks through the reasoning, the algorithm, and the implementation details, giving you a complete understanding of why the prefix‑sum method works and how to apply it efficiently.

Understanding the Problem

You are given an array nums of length n. In a single operation you may:

1. Choose an index i where nums[i] > 0.
2. Decrement nums[i] by 1.
3. Increment nums[(i + 1) % n] by 1.

The goal is to make every element of the array zero using the fewest possible operations.

Key constraints

• 1 ≤ n ≤ 10^5
• 0 ≤ nums[i] ≤ 10^9

Because the values can be huge, any solution that iterates over each unit individually will time out That's the part that actually makes a difference..

Core Insight – Moving Units Rightward

Think of each unit of value as a token that must travel to the right until it reaches the end of the array, where it “falls off” (since the last element’s right neighbor is the first element, but we never need to move tokens past the last index because we can always move them to the first index and later to the second, etc.).

When you decrement nums[i] and increment nums[i+1], you are effectively shifting one token one step to the right. The total number of shifts required for a token initially at position i to reach the end is equal to the sum of all values to its right.

If we denote the array as a[0], a[1], …, a[n‑1], the total number of moves needed is the sum of how many times each token must cross each boundary between adjacent indices. This can be expressed mathematically using prefix sums.

The Prefix‑Sum Formula

Define the prefix sum P[k] = a[0] + a[1] + … + a[k] for k = 0 … n‑1.
The minimum number of operations ops is:

ops = Σ_{k=0}^{n‑1} |P[k]|

Why does this work?

• P[k] represents the net amount of tokens that have passed the boundary between index k and k+1 after processing the first k+1 elements.
• If P[k] is positive, there are extra tokens that must move rightward across that boundary; each token contributes one operation per crossing.
• If P[k] is negative, it means we have more tokens arriving from the left than can stay, and we need to move tokens leftward (which is impossible directly). In the algorithm, negative values are handled by absorbing them into the next step, effectively counting the operations needed to bring the system back to zero.

Thus, taking the absolute value of each prefix sum and summing them yields the exact count of individual moves required Easy to understand, harder to ignore..

Step‑by‑Step Algorithm

1. Initialize prefix = 0 and operations = 0.
2. Iterate over each element num in nums:
   • Update prefix += num.
   • Add abs(prefix) to operations.
3. Return operations.

This single pass runs in linear time and uses constant extra space.

Example Walkthrough

Consider nums = [1, 2, 3].

| Index | num | prefix (running sum) | |prefix| added to ops | |-------|-----|----------------------|----------------------| | 0 | 1 | 1 | 1 | | 1 | 2 | 3 | 3 | | 2 | 3 | 6 | 6 |

Total operations = 1 + 3 + 6 = 10.

Verification – One possible sequence of operations (not minimal steps, just to see the pattern) could be:

• Move three tokens from index 2 rightward (they disappear after index 2).
• Move two tokens from index 1 rightward (they become index 2).
• Move one token from index 0 rightward (it becomes index 1).

Counting each individual shift yields exactly 10 moves, matching the formula Less friction, more output..

Implementation in Popular Languages

Python

def minimumOperations(nums):
    prefix = 0
    ops = 0
    for num in nums:
        prefix += num
        ops += abs(prefix)
    return ops

Java

class Solution {
    public long minimumOperations(int[] nums) {
        long prefix = 0;
        long ops = 0;
        for (int num : nums) {
            prefix += num;
            ops += Math.abs(prefix);
        }
        return ops;
    }
}

C++

long long minimumOperations(vector& nums) {
    long long prefix = 0, ops = 0;
    for (int num : nums) {
        prefix += num;
        ops += llabs(prefix);
    }
    return ops;
}

All three implementations follow the same linear‑time logic and handle large numbers using 64‑bit integers because the answer can exceed 2^31.

Complexity Analysis

• Time Complexity: O(n) – a single traversal of the array.
• Space Complexity: O(1) – only a few variables are used regardless of input size.

These bounds are optimal; any algorithm must at least read each element once.

Frequently Asked Questions

Q: Why do we need the absolute value?
A: The prefix sum can become negative when later elements are smaller than the accumulated total. A negative prefix indicates that we have “excess” tokens that must be moved leftward to balance the system, and each unit of imbalance still requires one operation to resolve Simple as that..

Q: Can we solve this using dynamic programming?
A: Yes, but DP would also be O(n) with higher constant overhead. The prefix‑sum method is the most concise and intuitive solution.

Q: Does the order of operations matter for the count?
A: No. Regardless of which index you choose at each step,

the total number of moves required to reach the zero state remains invariant. The prefix-sum formula captures this invariant directly, making the specific sequence of moves irrelevant to the final count The details matter here. Nothing fancy..

Q: What if the array contains only zeros? A: The algorithm correctly returns 0. The prefix sum remains zero throughout the iteration, adding nothing to the operation count.

Q: How does this relate to the "Minimum Operations to Make Array Equal" or similar problems? A: This specific problem—often titled "Minimum Operations to Make Array Zero" or "Minimum Moves to Balance Array"—is distinct because tokens can only move to adjacent indices (or fall off the ends). Problems allowing arbitrary redistribution usually reduce to calculating the median or mean, whereas this adjacent-move constraint leads directly to the prefix-sum solution.

Conclusion

The problem of minimizing operations to reduce an array to zero by shifting units between adjacent indices reveals a beautiful mathematical invariant: the total work required is exactly the sum of the absolute values of the running prefix sums.

This insight transforms a potentially complex simulation or dynamic programming challenge into a trivial single-pass algorithm. By recognizing that every unit of imbalance recorded in the prefix sum must cross the boundary to the next index, we derive an optimal O(n) time, O(1) space solution that is both elegant and highly performant. Whether implemented in Python, Java, C++, or any other language, this pattern serves as a powerful reminder that the most efficient algorithms often stem from a deep understanding of the problem's underlying physics rather than clever coding tricks.