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 eleement` ` ` `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 code` `int` `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 approach` `import` `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 eleement` ` ` `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 approach` ` ` `def` `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 eleement` ` ` `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 code` `if` `__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` |

**Output:**

2 0 4 7 8 6 5 1 3

**Time Complexity:** O(N)

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