Long Distance - Amazon Top Interview Questions

Problem Statement :

```Given a list of integers nums, return a new list where each element in the new list is the number of smaller elements to the right of that element in the original input list.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input

lst = [3, 4, 9, 6, 1]

Output

[1, 1, 2, 1, 0]

Explanation

There is 1 smaller element to the right of 3

There is 1 smaller element to the right of 4

There are 2 smaller elements to the right of 9

There is 1 smaller element to the right of 6

There are no smaller elements to the right of 1

Example 2

Input

lst = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Output

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Example 3

Input

lst = [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

Output

[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]```

Solution :

```                        ```Solution in C++ :

vector<int> ans;
void merge(vector<int> &v, int idx[], int l, int mid, int r) {
int l_n = mid - l + 1, r_n = r - mid;
int left[l_n], right[r_n];
for (int i = 0; i < l_n; i++) left[i] = idx[l + i];
for (int i = 0; i < r_n; i++) right[i] = idx[mid + 1 + i];
for (int i = l_n - 1, j = r_n - 1, k = r; k >= l; k--) {
if (j < 0 or (i >= 0 and v[left[i]] > v[right[j]])) {
ans[left[i]] += j + 1;
idx[k] = left[i--];
} else {
idx[k] = right[j--];
}
}
}

void mergesort(vector<int> &v, int idx[], int l, int r) {
if (l < r) {
int mid = l + (r - l) / 2;
mergesort(v, idx, l, mid);
mergesort(v, idx, mid + 1, r);
merge(v, idx, l, mid, r);
}
}

vector<int> solve(vector<int> &nums) {
int n = nums.size(), idx[n];
ans = vector<int>(n, 0);
for (int i = 0; i < n; i++) idx[i] = i;
mergesort(nums, idx, 0, n - 1);
return ans;
}```
```

```                        ```Solution in Java :

import java.util.*;

// This is esentially count the number of inversions problem.

class Solution {
public int[] solve(int[] lst) {
if (lst.length == 0)
return new int[0];

Point[] points = new Point[lst.length];

// convert given numbers to Points
for (int i = 0; i < lst.length; i++) {
points[i] = new Point(lst[i], i);
}

int[] result = new int[points.length];

mergeSort(points, 0, points.length - 1, result);

return result;
}

private Point[] mergeSort(Point[] points, int start, int end, int[] result) {
if (start >= end)
return new Point[] {points[start]};

int mid = start + (end - start) / 2;

Point[] left = mergeSort(points, start, mid, result);
Point[] right = mergeSort(points, mid + 1, end, result);

return merge(left, right, result);
}

private Point[] merge(Point[] left, Point[] right, int[] result) {
Point[] merged = new Point[left.length + right.length];

int i = 0;
int j = 0;

while (i < left.length && j < right.length) {
if (left[i].val <= right[j].val) {
merged[i + j] = left[i];
result[left[i].index] += j;
i++;
} else {
merged[i + j] = right[j];
j++;
}
}

while (i < left.length) {
merged[i + j] = left[i];
result[left[i].index] += j;
i++;
}

while (j < right.length) {
merged[i + j] = right[j];
j++;
}

return merged;
}

// Used as a utility class to easily identify the index of an element
private class Point {
int index;
int val;

public Point(int val, int i) {
this.val = val;
this.index = i;
}
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, lst):
def mergesort(indexes):
k = len(indexes)
if k <= 1:
return indexes
left = mergesort(indexes[: k // 2])
right = mergesort(indexes[k // 2 :])
for i in range(len(indexes) - 1, -1, -1):
if not right or left and lst[left[-1]] > lst[right[-1]]:
ans[left[-1]] += len(right)
indexes[i] = left.pop()
else:
indexes[i] = right.pop()
return indexes

n = len(lst)
ans = [0] * n
mergesort(list(range(n)))
return ans```
```

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