Count smaller elements present in the array for each array element
Given an array arr[] consisting of N integers, the task is for each array element, say arr[i], is to find the number of array elements that are smaller than arr[i].
Examples:
Input: arr[] = {3, 4, 1, 1, 2}
Output: 3 4 0 0 2
Explanation:
The elements which are smaller than arr[0](= 3) are {1, 1, 2}. Hence, the count is 3.
The elements which are smaller than arr[1](= 4) are {1, 1, 2, 3}. Hence, the count is 4.
The elements arr[2](= 1) and arr[3](= 1) are the smallest possible. Hence, the count is 0.
The elements which are smaller than arr[4](= 2) are {1, 1}. Hence, the count is 2.Input: arr[] = {1, 2, 3, 4}
Output: 0 1 2 3
Naive Approach: The simplest approach is to traverse the array and for each array element, count the number of array elements that are smaller than them and print the counts obtained.
Below is the implementation of the above approach:
- C++
- Java
- Python3
- C#
- Javascript
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count for each array// element, the number of elements// that are smaller than that elementvoid smallerNumbers(int arr[], int N){ // Traverse the array for (int i = 0; i < N; i++) { // Stores the count int count = 0; // Traverse the array for (int j = 0; j < N; j++) { // Increment count if (arr[j] < arr[i]) { count++; } } // Print the count of smaller // elements for the current element cout << count << " "; }}// Driver Codeint main(){ int arr[] = { 3, 4, 1, 1, 2 }; int N = sizeof(arr) / sizeof(arr[0]); smallerNumbers(arr, N); return 0;} |
Java
Python3
C#
Javascript
3 4 0 0 2
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by using Hashing. Follow the steps below to solve the problem:
- Initialize an auxiliary array hash[] of size 105 and initialize all array elements with 0 to store the frequency of each array element.
- Traverse the given array arr[] and increment the frequency of arr[i] by 1 in the array hash[] as hash[arr[i]]++.
- Find the prefix sum of the array hash[].
- Again, traverse the given array and for each array element arr[], print the value of hash[arr[i] – 1] as the count of smaller elements.
Below is the implementation of the above approach:
- C++
- Java
- Python3
- C#
- Javascript
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count for each array// element, the number of elements// that are smaller than that elementvoid smallerNumbers(int arr[], int N){ // Stores the frequencies // of array elements int hash[100000] = { 0 }; // Traverse the array for (int i = 0; i < N; i++) // Update frequency of arr[i] hash[arr[i]]++; // Initialize sum with 0 int sum = 0; // Compute prefix sum of the array hash[] for (int i = 1; i < 100000; i++) { hash[i] += hash[i - 1]; } // Traverse the array arr[] for (int i = 0; i < N; i++) { // If current element is 0 if (arr[i] == 0) { cout << "0"; continue; } // Print the resultant count cout << hash[arr[i] - 1] << " "; }}// Driver Codeint main(){ int arr[] = { 3, 4, 1, 1, 2 }; int N = sizeof(arr) / sizeof(arr[0]); smallerNumbers(arr, N); return 0;} |
Java
Python3
C#
Javascript
3 4 0 0 2
Time Complexity: O(N)
Auxiliary Space: O(N), where N is the maximum value of the array elements
