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Given an array with n distinct elements, convert the given array to a form where all elements are in range from 0 to n-1. The order of elements is same, i.e., 0 is placed in place of smallest element, 1 is placed for second smallest element, ... n-1 is placed for largest element.
Input: arr[] = {10, 40, 20}
Output: arr[] = {0, 2, 1}
Input: arr[] = {5, 10, 40, 30, 20}
Output: arr[] = {0, 1, 4, 3, 2}
We have discussed simple and hashing based solutions. In this post, a new solution is discussed. The idea is to create a vector of pairs. Every element of pair contains element and index. We sort vector by array values. After sorting, we copy indexes to original array.
Given Array is 10 20 15 12 11 50 Converted Array is 0 4 3 2 1 5
Time Complexity : O(n Log n)
Auxiliary Space : O(n)
Another Approach:
1) Create a copy of the input array and sort it in non-decreasing order.
2) Create a map where each element of the sorted array is mapped to its corresponding index in the range 0 to n-1.
3) Traverse the input array and for each element, replace it with the value obtained from the map.
Given Array is 10 20 15 12 11 50 Converted Array is 0 4 3 2 1 5
Time Complexity: O(n log n) where n is the size of the input array. This is because the function creates a copy of the input array using the "copy" function, which takes O(n) time, sorts the copy using the "sort" function, which has a time complexity of O(n log n)
Space Complexity: O(n) we are using extra space as a map.