Range Update - Google Top Interview Questions


Problem Statement :


You are given a list of integers nums and a two-dimensional list of integers operations.

 Each operation is of the following form: [L, R, X], which means that you should increment by X all the elements from indices L to R inclusive in the list (the list is 0-indexed).

Apply all operations and return the final list.

Constraints

n ≤ 10,000 where n is length of nums


o ≤ 10,000 where o is length of operations
Example 1

Input

nums = [7, 3, 1, -10, 3]

operations = [

    [0, 0, 3],

    [1, 3, 2],

    [2, 3, 5]

]

Output

[10, 5, 8, -3, 3]

Explanation

The initial list is [7, 3, 1, -10, 3].



After applying the first operation ([0, 0, 3]) the list becomes [10, 3, 1, -10, 3].

After applying the second operation ([1, 3, 2]) the list becomes [10, 5, 3, -8, 3].

After applying the third operation ([2, 3, 5]) the list becomes [10, 5, 8, -3, 3].


Solution :



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                        Solution in C++ :

vector<int> solve(vector<int>& nums, vector<vector<int>>& operations) {
    int n = nums.size();
    int increases[n + 1] = {0};
    for (vector<int>& op : operations) {
        int L = op[0];
        int R = op[1];
        int X = op[2];

        increases[L] += X;
        increases[R + 1] -= X;
    }

    int s = 0;
    vector<int> res;
    for (int i = 0; i < n; i++) {
        s += increases[i];
        res.push_back(s + nums[i]);
    }
    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int[] solve(int[] nums, int[][] operations) {
        if (nums == null || nums.length == 0)
            return nums;
        int[] temp = new int[nums.length + 1];
        for (int[] operation : operations) {
            int i = operation[0], j = operation[1], cost = operation[2];
            temp[i] += cost;
            temp[j + 1] -= cost;
        }

        for (int i = 1; i < temp.length; i++) temp[i] += temp[i - 1];

        for (int i = 0; i < nums.length; i++) nums[i] += temp[i];
        return nums;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums, operations):
        # [[0, 3], [1, -3], [1, 2], [2, 5], [4, -5], [4, -2]]
        # {0: 3, 1: -1, 2: 5, 4: -7}
        mp = [0 for _ in range(len(nums) + 1)]
        for op in operations:
            mp[op[0]] += op[2]
            mp[op[1] + 1] -= op[2]
        cur = 0
        for i in range(len(nums)):
            cur += mp[i]
            nums[i] += cur
        return nums


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