3356.zero-array-transformation-ii.cpp

//  Tag: Array, Binary Search, Prefix Sum, Line Sweep
//  Time: O(N + M)
//  Space: O(N)
//  Ref: -
//  Note: -
//  Video: https://youtu.be/h9SVvBXaVHY

//  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):
//  
//  Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.
//  The array will become [1, 0, 1].
//  
//  
//  For i = 1 (l = 0, r = 2, val = 1):
//  
//  Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.
//  The array will become [0, 0, 0], which is a Zero Array. Therefore, the minimum value of k is 2.
//  
//  
//  
//  
//  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):
//  
//  Decrement values at indices [1, 2, 3] by [2, 2, 1] respectively.
//  The array will become [4, 1, 0, 0].
//  
//  
//  For i = 1 (l = 0, r = 2, val = 1):
//  
//  Decrement values at indices [0, 1, 2] by [1, 1, 0] respectively.
//  The array will become [3, 0, 0, 0], which is not a Zero Array.
//  
//  
//  
//  
//   
//  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
//  
//  

class Solution {
public:
    int minZeroArray(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size();
        vector<int> line(n + 1, 0);
        int k = 0;
        int prefix = 0;
        for (int i = 0; i < n; i++) {
            while (prefix + line[i] < nums[i]) {
                if (k == queries.size()) {
                    return -1;
                }

                if (queries[k][1] >= i) {
                    line[max(i, queries[k][0])] += queries[k][2];
                    line[queries[k][1] + 1] -= queries[k][2];
                }

                k += 1;

            }

            prefix += line[i];
        }

        return k;
    }
};

class Solution {
public:
    int minZeroArray(vector<int>& nums, vector<vector<int>>& queries) {
        int n = queries.size();
        int l = 0;
        int r = n + 1;
        while (l < r) {
            int m = l + (r - l) / 2;
            if (fit(nums, queries, m)) {
                r = m;
            } else {
                l = m + 1;
            }
        }

        return l == n + 1 ? -1: l;
    }

    bool fit(vector<int>& nums, vector<vector<int>>& queries, int k) {
        int m = nums.size();
        vector<int> line(m + 1, 0);
        for (int i = 0; i < k; i++) {
            line[queries[i][0]] += queries[i][2];
            line[queries[i][1] + 1] -= queries[i][2];
        }

        int prefix = 0;
        for (int i = 0; i < m; i++) {
            prefix += line[i];
            if (nums[i] > prefix) {
                return false;
            }
        }

        return true;
    }
};