# Replace elements by its rank in the array

Replace elements by its rank in the array

Problem Statement: Given an array of N integers, the task is to replace each element of the array by its rank in the array.

Examples:

```Example 1:
Input: 20 15 26 2 98 6
Output: 4 3 5 1 6 2
Explanation: When sorted,the array is 2,6,15,20,26,98. So the rank of 2 is 1,rank of 6 is 2,rank of 15 is 3 and so…

Example 2:
Input: 1 5 8 15 8 25 9
Output: 1 2 3 5 3 6 4
Explanation: When sorted,the array is 1,5,8,8,9,15,25. So the rank of 1 is 1,rank of 5 is 2,rank of 8 is 3 and so…```

Solution 1:Naive approach

Intuition: Use two for loops and calculate a rank for each element.

Approach:

• Iterative over the array using a for loop.
• Now use another for loop to check the number of elements having value less than the current element.
• We can use a set to get the number of unique elements having value less than the current element.
• Number of such unique elements + 1 is the rank of that element.

Code:

## C++ Code

``````#include<bits/stdc++.h>
using namespace std;
int main()
{
int n = 6;
int arr[n] = {20, 15, 26, 2, 98, 6};
for (int i = 0; i < n; i++) {
set<int>s;
for (int j = 0; j < n; j++) {
if (arr[j] < arr[i]) {
s.insert(arr[j]);
}
}
cout << s.size() + 1 << " ";
}

}
``````

Output: 4 3 5 1 6 2

Time Complexity: O(N*N) as we are using double for loop.

Space Complexity: O(N) as we are using set.

## Java Code

``````import java.util.*;
public class Main {
public static void main(String args[]) {
int n = 6;
int arr[] = {20,15,26,2,98,6};
for (int i = 0; i < n; i++) {
Set < Integer > s = new HashSet < Integer > ();
for (int j = 0; j < n; j++) {
if (arr[j] < arr[i]) {
}
}
int rank = s.size() + 1;
System.out.print(rank + " ");
}
}
}
``````

Output: 4 3 5 1 6 2

Time Complexity: O(N*N) as we are using double for loop.

Space Complexity: O(N) as we are using set.

Solution 2: optimized

Intuition: Sort the array and store the rank of the element in a map.

Approach:

• Copy all the elements of the array in a dummy array.
• Sort the dummy array.
• Take a variable temp to calculate the rank of each element.Initially the value of temp is 1.
• Maintain a map to store the rank of each element.
• Now traverse through the dummy array,if element is previously not stored in map then store the value of element with the temp and then increament temp.
• Now traverse through the original array and print the rank of that corresponding element using map.

Code:

## C++ Code

``````#include<bits/stdc++.h>
using namespace std;
int main()
{
int n = 6;
int arr[n] = {20, 15, 26, 2, 98, 6};
map<int, int>mp;
int temp = 1;
int brr[n];
for (int i = 0; i < n; i++) {
brr[i] = arr[i];
}
sort(brr, brr + n);
for (int i = 0; i < n; i++) {
//if element is previously not store in the map
if (mp[brr[i]] == 0) {
mp[brr[i]] = temp;
temp++;
}
}
for (int i = 0; i < n; i++) {
cout << mp[arr[i]] << " ";
}

}
``````

Output: 4 3 5 1 6 2

Time Complexity:O(n)+O(nlogn)+O(n). O(n) for copying the elements from the original to the dummy array, O(nlogn) for sorting the dummy array, and O(n) for printing the ranks.

Space Complexity:O(N+N), for using dummy array and hashmap.

## Java Code

``````import java.util.*;
public class Main {
public static void main(String args[]) {
int arr[] = {20,15,26,2,98,6};
HashMap < Integer, Integer > mp = new HashMap < Integer, Integer > ();
int temp = 1;
int n = arr.length;
int brr[] = new int[n];
for (int i = 0; i < n; i++) {
brr[i] = arr[i];
}
Arrays.sort(brr);
for (int i = 0; i < n; i++) {
//if element is previously not store in the map
if (mp.get(brr[i]) == null) {
mp.put(brr[i], temp);
temp++;
}
}
for (int i = 0; i < n; i++) {
System.out.print(mp.get(arr[i]) + " ");
}
}
}
``````

Output: 4 3 5 1 6 2

Time Complexity:O(n)+O(nlogn)+O(n). O(n) for copying the elements from the original to the dummy array, O(nlogn) for sorting the dummy array, and O(n) for printing the ranks.

Space Complexity:O(N+N), for using dummy array and hashmap.

Special thanks to Pranav Padawe for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article