2025-03-13 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's March LeetCoding Challenge with cpp.

March LeetCoding Challenge 13

Description

Zero Array Transformation II

You are given an integer array nums of length n and a 2D array queries where queries[i] = [li, ri, vali].

Each queries[i] represents the following action on nums:

  • Decrement the value at each index in the range [li, ri] in nums by at most vali.
  • The amount by which each value is decremented can be chosen independently for each index.

A Zero Array is an array with all its elements equal to 0.

Return the minimum possible non-negative value of k, such that after processing the first k queries in sequence, nums becomes a Zero Array. If no such k exists, return -1.

 

Example 1:

Input: nums = [2,0,2], queries = [[0,2,1],[0,2,1],[1,1,3]]

Output: 2

Explanation:

  • For i = 0 (l = 0, r = 2, val = 1): 
    <ul>
	<li>Decrement values at indices <code>[0, 1, 2]</code> by <code>[1, 0, 1]</code> respectively.</li>
	<li>The array will become <code>[1, 0, 1]</code>.</li>
</ul>
</li>
<li><strong>For i = 1 (l = 0, r = 2, val = 1):</strong>
<ul>
	<li>Decrement values at indices <code>[0, 1, 2]</code> by <code>[1, 0, 1]</code> respectively.</li>
	<li>The array will become <code>[0, 0, 0]</code>, which is a Zero Array. Therefore, the minimum value of <code>k</code> is 2.</li>
</ul>
</li>

Example 2:

Input: nums = [4,3,2,1], queries = [[1,3,2],[0,2,1]]

Output: -1

Explanation:

  • For i = 0 (l = 1, r = 3, val = 2): 
    <ul>
	<li>Decrement values at indices <code>[1, 2, 3]</code> by <code>[2, 2, 1]</code> respectively.</li>
	<li>The array will become <code>[4, 1, 0, 0]</code>.</li>
</ul>
</li>
<li><strong>For i = 1 (l = 0, r = 2, val<span style="font-size: 13.3333px;"> </span>= 1):</strong>
<ul>
	<li>Decrement values at indices <code>[0, 1, 2]</code> by <code>[1, 1, 0]</code> respectively.</li>
	<li>The array will become <code>[3, 0, 0, 0]</code>, which is not a Zero Array.</li>
</ul>
</li>

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 5 * 105
  • 1 <= queries.length <= 105
  • queries[i].length == 3
  • 0 <= li <= ri < nums.length
  • 1 <= vali <= 5

Solution

class Solution {
public:
  int minZeroArray(vector<int>& nums, vector<vector<int>>& queries) {
    int decrement = 0;
    int k = 0;
    vector<int> decrements(nums.size() + 1);
    for(int i = 0; i < nums.size(); ++i) {
      for(; decrement < nums[i] - decrements[i]; ++k) {
        if(k >= queries.size()) return -1;
        if(queries[k][1] < i) continue;
        decrements[max(queries[k][0], i)] += queries[k][2];
        decrements[queries[k][1] + 1] -= queries[k][2];
      }
      decrement += decrements[i];
    }
    return k;
  }
};

// Accepted
// 627/627 cases passed (8 ms)
// Your runtime beats 82.19 % of cpp submissions
// Your memory usage beats 74.49 % of cpp submissions (323 MB)
Algorithm