Given an array arr of integers of size N, the task is to find, for every element, the number of elements that are greater than it.
Examples:
Input: arr[] = {4, 6, 2, 1, 8, 7}
Output: {3, 2, 4, 5, 0, 1}
Input: arr[] = {2, 3, 4, 5, 6, 7, 8}
Output: {6, 5, 4, 3, 2, 1, 0}
Approach: Store the frequencies of every array element using a Map. Iterate the Map in reverse and store the sum of the frequency of all previously traversed elements (i.e. elements greater than it) for every element.
Below code is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;void countOfGreaterElements(int arr[], int n){ // Store the frequency of the // array elements map<int, int> mp; for (int i = 0; i < n; i++) { mp[arr[i]]++; } int x = 0; // Store the sum of frequency of elements // greater than the current element for (auto it = mp.rbegin(); it != mp.rend(); it++) { int temp = it->second; mp[it->first] = x; x += temp; } for (int i = 0; i < n; i++) cout << mp[arr[i]] << " ";}// Driver codeint main(){ int arr[] = { 7, 9, 5, 2, 1, 3, 4, 8, 6 }; int n = sizeof(arr) / sizeof(arr[0]); countOfGreaterElements(arr, n); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GfG { public static void countOfGreaterElements(int arr[]) { int n = arr.length; TreeMap<Integer, Integer> mp = new TreeMap<Integer, Integer>(Collections.reverseOrder()); // Store the frequency of the // array elements for (int i = 0; i < n; i++) { mp.put(arr[i], mp.getOrDefault(arr[i], 0) + 1); } // Store the sum of frequency of elements // greater than the current element int x = 0; for (Map.Entry<Integer, Integer> e : mp.entrySet()) { Integer temp = e.getValue(); mp.put(e.getKey(), x); x += temp; } for (int i = 0; i < n; i++) System.out.print(mp.get(arr[i]) + " "); } public static void main(String args[]) { int arr[] = { 7, 9, 5, 2, 1, 3, 4, 8, 6 }; countOfGreaterElements(arr); }} |
Python 3
# Python 3 implementation of the above approachdef countOfGreaterElements(arr, n): # Store the frequency of the # array elements mp = {i:0 for i in range(1000)} for i in range(n): mp[arr[i]] += 1 x = 0 # Store the sum of frequency of elements # greater than the current element p = [] q = [] m = [] for key, value in mp.items(): m.append([key, value]) m = m[::-1] for p in m: temp = p[1] mp[p[0]] = x x += temp for i in range(n): print(mp[arr[i]], end = " ")# Driver codeif __name__ == '__main__': arr = [7, 9, 5, 2, 1, 3, 4, 8, 6] n = len(arr) countOfGreaterElements(arr, n)# This code is contributed by Surendra_Gangwar |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic;class GfG { public static void countOfGreaterElements(int[] arr) { int n = arr.Length; SortedDictionary<int, int> mp = new SortedDictionary<int, int>(); // Store the frequency of the // array elements for (int i = 0; i < n; i++) { int temp = 0; if (mp.ContainsKey(arr[i])) { temp = mp[arr[i]]; mp.Remove(arr[i]); } mp[arr[i]] = temp + 1; } // Store the sum of frequency of elements // greater than the current element int x = 0; var k1 = mp.Keys; int[] keys = new int[mp.Count]; k1.CopyTo(keys, 0); Array.Reverse(keys); foreach (var e in keys) { int temp = mp[e]; mp.Remove(e); mp[e] = x; x += temp; } for (int i = 0; i < n; i++) Console.Write(mp[arr[i]] + " "); } public static void Main(string[] args) { int[] arr = { 7, 9, 5, 2, 1, 3, 4, 8, 6 }; countOfGreaterElements(arr); }}// This code is contributed by phasing17 |
Javascript
// JavaScript implementation of the above approachfunction countOfGreaterElements(arr, n){ // Store the frequency of the // array elements let mp = {} for (var i = 0; i < n; i++) { if (!mp.hasOwnProperty(arr[i])) mp[arr[i]] = 0 mp[arr[i]] = mp[arr[i]] + 1 } let x = 0 // Store the sum of frequency of elements // greater than the current element let m = [] for (var [key, value] of Object.entries(mp)) { key = parseInt(key) value = parseInt(value) m.push([key, value]) } for (var i = m.length - 1; i >= 0; i--) { var p = m[i] var temp = p[1] mp[p[0]] = x x += temp } for (var i = 0; i < n; i++) process.stdout.write(mp[arr[i]] + " ")}// Driver codelet arr = [7, 9, 5, 2, 1, 3, 4, 8, 6]let n = arr.lengthcountOfGreaterElements(arr, n)// This code is contributed by phasing17 |
Output:
2 0 4 7 8 6 5 1 3
Time Complexity: O(N log(N))
Auxiliary Space: O(N)
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