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 :
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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