Leetcode 1658: Minimum Operations to Reduce X to Zero

Input: You are given two inputs: an integer array `nums` and an integer `x`.

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

Constraints:

• 1 <= nums.length <= 10^5

• 1 <= nums[i] <= 10^4

• 1 <= x <= 10^9

Output: Return the minimum number of operations needed to reduce `x` to 0, or -1 if it's not possible.

Example: 2

Constraints:

• The output will be a single integer: either the minimum number of operations or -1.

Goal: Determine the minimum number of operations to reduce `x` to 0, by removing elements from either end of the array and subtracting their value from `x`.

Steps:

• Calculate the total sum of elements in `nums`. If the sum is smaller than `x`, return -1 immediately.

• Use a sliding window approach to check the sum of subarrays of `nums` that can contribute to reducing `x` to 0.

• Keep track of the longest subarray that matches the condition of reducing `x` to 0, and calculate the number of operations.

Goal: The constraints are designed to ensure the solution can handle large arrays and large values of `x` efficiently.

Steps:

• 1 <= nums.length <= 10^5

• 1 <= nums[i] <= 10^4

• 1 <= x <= 10^9

Assumptions:

• The array `nums` contains only positive integers.

• The array has a length between 1 and 100,000, and `x` can be as large as 1 billion.

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

• Explanation: In this case, we can remove the last two elements (4 + 1 = 5) to reduce `x` to 0, requiring exactly 2 operations.

• Input: nums = [4, 5, 6, 7, 8], x = 3

• Explanation: In this case, it’s impossible to remove elements from the array to sum to `x = 3`. Hence, the answer is -1.

Approach: We will use a sliding window approach to find the longest subarray that can reduce `x` to 0 by removing elements from the array.

Observations:

• The problem can be reduced to finding a subarray whose sum is equal to `sum(nums) - x`.

• Using a sliding window or prefix sum technique will help optimize the solution for large inputs.

• Start by checking if the sum of all elements in `nums` is smaller than `x`. If it is, return -1 immediately.

• Otherwise, calculate the difference between the sum of the array and `x`.

Steps:

• Compute the total sum of elements in `nums`.

• Calculate the target sum as `total_sum - x`.

• Use a sliding window or prefix sum approach to find the longest subarray with the target sum.

• If such a subarray is found, the number of operations is `n - length of the subarray`. If no such subarray exists, return -1.

Empty Inputs:

• If `nums` is empty, return -1 immediately.

Large Inputs:

• For large arrays, ensure that the algorithm handles the input within acceptable time limits.

Special Values:

• If `x` is greater than the sum of all elements in `nums`, return -1.

Constraints:

• Ensure that the solution works efficiently with up to 100,000 elements.

int minOperations(vector<int>& nums, int x) {
    int res = -1;
    long sum = -x;
    for(int y: nums) sum += y;
    int n = nums.size();
    if (sum == 0) return n;
    
    map<int, int> mp;
    mp[0] = -1;
    int s = 0;
    for(int i = 0; i < n; i++) {
        s += nums[i];
        if(mp.count(s - sum)) {
            res = max(res, i - mp[s - sum]);
        }
        mp[s] = i;
    }
    
    return res == -1? res :n - res;        
}

1 : Function Definition

int minOperations(vector<int>& nums, int x) {

Defines the function `minOperations` which takes an array and a target value as inputs.

2 : Variable Initialization

    int res = -1;

Initializes the result variable `res` to -1 to track the maximum subarray length.

3 : Variable Initialization

    long sum = -x;

Initializes `sum` to -x to adjust the target for finding a subarray with a specific sum.

4 : Looping

    for(int y: nums) sum += y;

Calculates the total sum of the array and adjusts `sum` for further processing.

5 : Variable Initialization

    int n = nums.size();

Stores the size of the array in `n` for use in subsequent logic.

6 : Condition Check

    if (sum == 0) return n;

Checks if the total sum equals `x`. If true, the entire array is the solution.

7 : Map Initialization

    map<int, int> mp;

Initializes a map to store prefix sums and their corresponding indices.

8 : Map Update

    mp[0] = -1;

Adds an initial entry to the map for handling prefix sums starting from the beginning.

9 : Variable Initialization

    int s = 0;

Initializes the prefix sum variable `s` to zero.

10 : Looping

    for(int i = 0; i < n; i++) {

Iterates through the array to compute prefix sums and check for subarray matches.

11 : Sum Update

        s += nums[i];

Updates the prefix sum `s` with the current element.

12 : Condition Check

        if(mp.count(s - sum)) {

Checks if there exists a prefix sum in the map such that the difference equals the target sum.

13 : Update Result

            res = max(res, i - mp[s - sum]);

Updates `res` to the maximum length of a valid subarray found so far.

14 : Map Update

        mp[s] = i;

Adds the current prefix sum and its index to the map for future reference.

15 : Return

    return res == -1? res :n - res;

Returns the minimum operations needed by converting the subarray length to elements removed.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is linear because we scan the array once using the sliding window technique.

Best Case: O(1)

Worst Case: O(n)

Description: The space complexity is O(n) due to the storage of intermediate results like prefix sums or hash maps.

Leetcode 1658: Minimum Operations to Reduce X to Zero

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