What is the time complexity of Minimum Operations to Reduce X to Zero?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Operations to Reduce X to Zero?

This problem uses Array, Hash Table, Binary Search, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Operations to Reduce X to Zero?

Insight 1: Instead of searching for the minimum number of operations to reduce `x` to zero by removing elements from both ends, reframe the problem. Find the longest subarray whose sum equals `total_sum - x`. The remaining elements outside this subarray will represent the elements removed from both ends. Insight 2: Use a sliding window approach to efficiently find the longest subarray with the required sum.

Which companies ask LeetCode 1658. Minimum Operations to Reduce X to Zero in interviews?

This problem is frequently asked at Morgan Stanley.

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Minimum Operations to Reduce X to Zero - LeetCode 1658 Solution

Minimum Operations to Reduce X to Zero - Complete Solution Guide

Minimum Operations to Reduce X to Zero is LeetCode problem 1658, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer array nums and an integer x . In one operation, you can either remove the leftmost or the rightmost element from the array nums and subtract its value from x . Note that this modifies the array for future operations. Return the minimum number of operations to reduce x to exactly 0 if it is possible , otherwise, return -1 . Example 1: Input: nums = [1,1,4,2,3], x = 5 Output: 2 Explanation: The optimal solution is to remove the last two elements to reduce x to zero. Exampl

Detailed Explanation

The problem requires finding the minimum number of operations to reduce an integer `x` to zero by removing elements from either the leftmost or rightmost end of an integer array `nums`. Each removal subtracts the element's value from `x`. The array is modified after each removal. The goal is to return the minimum number of removals (operations) needed, or -1 if it's impossible to reduce `x` to zero.

Solution Approach

The solution uses a sliding window technique to find the longest contiguous subarray within `nums` whose sum equals `total_sum - x`. First, it calculates the total sum of `nums`. Then, it determines the target sum by subtracting `x` from the total sum. A sliding window is maintained using `left` and `right` pointers. The window expands by including elements from the right, and contracts by removing elements from the left until the window's sum is less than or equal to the target sum. If the window's sum equals the target sum, the length of the window is compared with the maximum length found so far. Finally, the minimum number of operations is computed as the original array length minus the maximum length of the subarray with the target sum.

Step-by-Step Algorithm

Step 1: Calculate the total sum of all elements in the `nums` array.
Step 2: Calculate the `target` sum by subtracting `x` from the `total_sum`. This `target` sum represents the sum of the longest subarray we need to find.
Step 3: Handle edge cases: If `target` is negative, it's impossible to reduce `x` to zero, so return -1. If `target` is 0, the whole array sums to x, so we return the array length.
Step 4: Initialize `max_len` to -1, `current_sum` to 0, and `left` pointer to 0.
Step 5: Iterate through the array using the `right` pointer from 0 to `n-1`.
Step 6: Add the current element `nums[right]` to `current_sum`.
Step 7: While `current_sum` is greater than `target` and `left` is less than or equal to `right`, subtract `nums[left]` from `current_sum` and increment `left` to shrink the window from the left.
Step 8: If `current_sum` is equal to `target`, update `max_len` with the maximum of its current value and `right - left + 1` (the length of the current window).
Step 9: After the loop finishes, if `max_len` is still -1, it means no subarray with the target sum was found, so return -1. Otherwise, return `n - max_len` (the minimum number of operations).

Key Insights

Insight 1: Instead of searching for the minimum number of operations to reduce `x` to zero by removing elements from both ends, reframe the problem. Find the longest subarray whose sum equals `total_sum - x`. The remaining elements outside this subarray will represent the elements removed from both ends.
Insight 2: Use a sliding window approach to efficiently find the longest subarray with the required sum.
Insight 3: Handle the edge cases where the target sum (total_sum - x) is negative, zero, or when no such subarray exists.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Binary Search, Sliding Window, Prefix Sum.

Companies

Asked at: Morgan Stanley.

Minimum Operations to Reduce X to Zero | LeetCode 1658

Medium

Array Hash Table Binary Search Sliding Window Prefix Sum

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minOperations(int[] nums, int x) {
        int total_sum = 0;
        for (int num : nums) {
            total_sum += num;
        }
        int target = total_sum - x;
        int n = nums.length;

        if (target < 0) {
            return -1;
        }
        
        if (target == 0) {
            return n;
        }

        int max_len = -1;
        int current_sum = 0;
        int left = 0;
        
        for (int right = 0; right < n; right++) {
            current_sum += nums[right];
            
            while (current_sum > target && left <= right) {
                current_sum -= nums[left];
                left++;
            }
            
            if (current_sum == target) {
                max_len = Math.max(max_len, right - left + 1);
            }
        }

        if (max_len == -1) {
            return -1;
        } else {
            return n - max_len;
        }
    }
}

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Problem Info

Description

Problem 1658: Minimum Operations to Reduce X to Zero - Detailed explanation and solution approaches.

Topics

Array Hash Table Binary Search Sliding Window Prefix Sum

Companies

Morgan Stanley

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1658

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Calculate the total sum of all elements in the `nums` array.
2
Step 2: Calculate the `target` sum by subtracting `x` from the `total_sum`. This `target` sum represents the sum of the longest subarray we need to find.
3
Step 3: Handle edge cases: If `target` is negative, it's impossible to reduce `x` to zero, so return -1. If `target` is 0, the whole array sums to x, so we return the array length.
4
Step 4: Initialize `max_len` to -1, `current_sum` to 0, and `left` pointer to 0.
5
Step 5: Iterate through the array using the `right` pointer from 0 to `n-1`.
6
Step 6: Add the current element `nums[right]` to `current_sum`.
7
Step 7: While `current_sum` is greater than `target` and `left` is less than or equal to `right`, subtract `nums[left]` from `current_sum` and increment `left` to shrink the window from the left.
8
Step 8: If `current_sum` is equal to `target`, update `max_len` with the maximum of its current value and `right - left + 1` (the length of the current window).
9
Step 9: After the loop finishes, if `max_len` is still -1, it means no subarray with the target sum was found, so return -1. Otherwise, return `n - max_len` (the minimum number of operations).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because each element of the array is visited at most twice (once by the `right` pointer and at most once by the `left` pointer). The space complexity is O(1) because only a few integer variables are used, regardless of the size of the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Prefix Sum and Hash Table

Compute prefix sums of the array and store them in a hash table. Then, iterate through the prefix sums and check if `prefix_sum[i] - target` exists in the hash table. The difference between the indices can then be used to calculate the length of the subarray.

Time: O(n)Space: O(n)

Trade-offs: The prefix sum and hash table approach also has a time complexity of O(n), but it uses O(n) space. The sliding window approach is generally preferred due to its constant space complexity. However, if the problem had additional constraints or requirements, the hash table approach might be more suitable.

✨

Key Insights

1Insight 1: Instead of searching for the minimum number of operations to reduce `x` to zero by removing elements from both ends, reframe the problem. Find the longest subarray whose sum equals `total_sum - x`. The remaining elements outside this subarray will represent the elements removed from both ends.
2Insight 2: Use a sliding window approach to efficiently find the longest subarray with the required sum.
3Insight 3: Handle the edge cases where the target sum (total_sum - x) is negative, zero, or when no such subarray exists.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: `target` is negative (i.e., `total_sum - x < 0`). This means it's impossible to reduce `x` to zero, so the function correctly returns -1.
Edge case 2: `target` is zero (i.e., `total_sum == x`). This means removing no element leads to x = 0, and the whole array's sum is equal to x, hence the function correctly returns the array length `n`.
Edge case 3: No subarray with the sum equal to the target sum is found. In this case, `max_len` remains -1, and the function correctly returns -1.
Edge case 4: nums is null or empty. The provided code implicitly handles an empty array since the initial sum is zero. If `x` is also zero, it returns 0. Otherwise it will result in target being negative which returns -1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case where the target sum is negative or 0. This will lead to incorrect results.

✗Common mistake 2: Not handling the case where no subarray with the required sum exists. This will lead to returning an incorrect number of operations.

✗Common mistake 3: Forgetting to update `max_len` when a subarray with the target sum is found. Initializing max_len to 0 instead of -1 could also be an issue if no such array exists.

✗Common mistake 4: Incorrectly shrinking the window, potentially skipping valid subarrays.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation - Optimizing the usage of resources by identifying the longest period where resource consumption matches a specific target. Removing resources from the start or end represent allocating to other projects.

🏦Real-world scenario 2: Financial analysis - Identifying the longest period of stock prices that sum up to a certain target profit or loss. The removals from beginning or end represents trimming the period and calculating return from the remaining segment.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Operations to Reduce X to Zero"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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