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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