# 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```
```

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink