Related Articles

# Making elements distinct in a sorted array by minimum increments

• Difficulty Level : Medium
• Last Updated : 03 May, 2021

Given a sorted integer array. We need to make array elements distinct by increasing values and keeping array sum minimum possible. We need to print the minimum possible sum as output.

Examples:

```Input : arr[] = { 2, 2, 3, 5, 6 } ;
Output : 20
Explanation : We make the array as {2,
3, 4, 5, 6}. Sum becomes 2 + 3 + 4 +
5 + 6 = 20

Input : arr[] = { 20, 20 } ;
Output : 41
Explanation : We make {20, 21}

Input :  arr[] = { 3, 4, 6, 8 };
Output : 21
Explanation : All elements are unique
so result is sum of each elements.
```

Method 1:
1. Traverse each element of array .
2. if arr[i] == arr[i-1] then update each element of array by adding 1 from i-th(current) position to where element is either equal to its previous element or has become less than previous (because previous was increased).
3. After traversing of each element return sum .

## C++

 `// CPP program to make sorted array elements``// distinct by incrementing elements and keeping``// sum to minimum.``#include ``using` `namespace` `std;` `// To find minimum sum of unique elements.``int` `minSum(``int` `arr[], ``int` `n)``{``    ``int` `sum = arr;` `    ``for` `(``int` `i = 1; i < n; i++) {``        ``if` `(arr[i] == arr[i - 1]) {           ` `            ``// While current element is same as``            ``// previous or has become smaller``            ``// than previous.``            ``int` `j = i;``            ``while` `(j < n && arr[j] <= arr[j - 1]) {         ``                ``arr[j] = arr[j] + 1;``                ``j++;``            ``}``        ``}``         ``sum = sum + arr[i];``     ``}` `    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 2, 3, 5, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << minSum(arr, n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to make sorted``// array elements distinct by``// incrementing elements and``// keeping sum to minimum.``import` `java.io.*;` `class` `GFG``{``    ``// To find minimum sum``    ``// of unique elements.``    ``static` `int` `minSum(``int` `arr[], ``int` `n)``    ``{``        ``int` `sum = arr[``0``];``    ` `        ``for` `(``int` `i = ``1``; i < n; i++)``        ``{``            ``if` `(arr[i] == arr[i - ``1``]) {        ``    ` `                ``// While current element is same as``                ``// previous or has become smaller``                ``// than previous.``                ``int` `j = i;``                ``while` `(j < n && arr[j] <= arr[j - ``1``])``                ``{        ``                    ``arr[j] = arr[j] + ``1``;``                    ``j++;``                ``}``            ``}``            ``sum = sum + arr[i];``        ``}``    ` `        ``return` `sum;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = { ``2``, ``2``, ``3``, ``5``, ``6` `};``        ``int` `n = arr.length;``        ``System.out.println(minSum(arr, n));``    ``}``}` `// This code is contributed by Ansu Kumari`

## Python3

 `# Python3 program to make sorted array elements``# distinct by incrementing elements and keeping``# sum to minimum.` `# To find minimum sum of unique elements.``def` `minSum(arr, n):``    ``sm ``=` `arr[``0``]` `    ``for` `i ``in` `range``(``1``, n):``        ``if` `arr[i] ``=``=` `arr[i ``-` `1``]:        ` `            ``# While current element is same as``            ``# previous or has become smaller``            ``# than previous.``            ``j ``=` `i``            ``while` `j < n ``and` `arr[j] <``=` `arr[j ``-` `1``]:        ``                ``arr[j] ``=` `arr[j] ``+` `1``                ``j ``+``=` `1` `        ``sm ``=` `sm ``+` `arr[i]` `    ``return` `sm` `# Driver code``arr ``=` `[ ``2``, ``2``, ``3``, ``5``, ``6` `]``n ``=` `len``(arr)``print``(minSum(arr, n))` `# This code is contributed by Ansu Kumari`

## C#

 `// C# program to make sorted``// array elements distinct by``// incrementing elements and``// keeping sum to minimum.``using` `System;` `class` `GFG``{``    ``// To find minimum sum``    ``// of unique elements.``    ``static` `int` `minSum(``int` `[]arr, ``int` `n)``    ``{``        ``int` `sum = arr;``    ` `        ``for` `(``int` `i = 1; i < n; i++)``        ``{``            ``if` `(arr[i] == arr[i - 1]) {    ``    ` `                ``// While current element is same as``                ``// previous or has become smaller``                ``// than previous.``                ``int` `j = i;``                ``while` `(j < n && arr[j] <= arr[j - 1])``                ``{    ``                    ``arr[j] = arr[j] + 1;``                    ``j++;``                ``}``            ``}``            ``sum = sum + arr[i];``        ``}``    ` `        ``return` `sum;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = { 2, 2, 3, 5, 6 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(minSum(arr, n));``    ``}``}` `// This code is contributed by vt_m`

## PHP

 ``

## Javascript

 ``

Output :

`20`

Time Complexity : O(n^2)

Method 2:
1. Traverse each element of array .
2. If arr[i] <= prev then update prev by adding 1 and update sum by adding prev,
else update prev by cur element and update sum by adding cur element(arr[i]).
3. After traversing of each element return sum .

## C++

 `// Efficient CPP program to make sorted array``// elements distinct by incrementing elements``// and keeping sum to minimum.``#include ``using` `namespace` `std;` `// To find minimum sum of unique elements.``int` `minSum(``int` `arr[], ``int` `n)``{``    ``int` `sum = arr, prev = arr;` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// If violation happens, make current``        ``// value as 1 plus previous value and``        ``// add to sum.``        ``if` `(arr[i] <= prev) {``            ``prev = prev + 1;``            ``sum = sum + prev;``        ``}` `        ``// No violation.``        ``else` `{``            ``sum = sum + arr[i];``            ``prev = arr[i];``        ``}``    ``}` `    ``return` `sum;``}` `// Drivers code``int` `main()``{``    ``int` `arr[] = { 2, 2, 3, 5, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << minSum(arr, n) << endl;``    ``return` `0;``}`

## Java

 `// Efficient Java program to make sorted array``// elements distinct by incrementing elements``// and keeping sum to minimum.``import` `java.io.*;` `class` `GFG {` `    ``// To find minimum sum of unique elements.``    ``static` `int` `minSum(``int` `arr[], ``int` `n)``    ``{``        ``int` `sum = arr[``0``], prev = arr[``0``];``    ` `        ``for` `(``int` `i = ``1``; i < n; i++) {``    ` `            ``// If violation happens, make current``            ``// value as 1 plus previous value and``            ``// add to sum.``            ``if` `(arr[i] <= prev) {``                ``prev = prev + ``1``;``                ``sum = sum + prev;``            ``}``    ` `            ``// No violation.``            ` `            ``else` `{``                ``sum = sum + arr[i];``                ``prev = arr[i];``            ``}``        ``}``    ` `        ``return` `sum;``    ``}``    ` `    ``// Drivers code``    ``public` `static` `void` `main (String[] args) {``    ` `        ``int` `arr[] = { ``2``, ``2``, ``3``, ``5``, ``6` `};``        ``int` `n = arr.length;``        ` `        ``System.out.println(minSum(arr, n));``    ``}``}` `// This code is contributed by Ansu Kumari.`

## Python3

 `# Efficient Python program to make sorted array``# elements distinct by incrementing elements``# and keeping sum to minimum.` `# To find minimum sum of unique elements``def` `minSum(arr, n):``    ` `    ``sum` `=` `arr[``0``]; prev ``=` `arr[``0``]` `    ``for` `i ``in` `range``(``1``, n):` `        ``# If violation happens, make current``        ``# value as 1 plus previous value and``        ``# add to sum.``        ``if` `arr[i] <``=` `prev:``            ``prev ``=` `prev ``+` `1``            ``sum` `=` `sum` `+` `prev` `        ``# No violation.``        ``else` `:``            ``sum` `=` `sum` `+` `arr[i]``            ``prev ``=` `arr[i]` `    ``return` `sum` `# Drivers code``arr ``=` `[ ``2``, ``2``, ``3``, ``5``, ``6` `]``n ``=` `len``(arr)``print``(minSum(arr, n))` `# This code is contributed by Ansu Kumari`

## C#

 `// Efficient C# program to make sorted array``// elements distinct by incrementing elements``// and keeping sum to minimum.``using` `System;` `class` `GFG {` `    ``// To find minimum sum of unique elements.``    ``static` `int` `minSum(``int` `[]arr, ``int` `n)``    ``{``        ``int` `sum = arr, prev = arr;``    ` `        ``for` `(``int` `i = 1; i < n; i++) {``    ` `            ``// If violation happens, make current``            ``// value as 1 plus previous value and``            ``// add to sum.``            ``if` `(arr[i] <= prev) {``                ``prev = prev + 1;``                ``sum = sum + prev;``            ``}``    ` `            ``// No violation.``            ` `            ``else` `{``                ``sum = sum + arr[i];``                ``prev = arr[i];``            ``}``        ``}``    ` `        ``return` `sum;``    ``}``    ` `    ``// Drivers code``    ``public` `static` `void` `Main () {``    ` `        ``int` `[]arr = { 2, 2, 3, 5, 6 };``        ``int` `n = arr.Length;``        ` `        ``Console.WriteLine(minSum(arr, n));``    ``}``}` `// This code is contributed by vt_m .`

## PHP

 ``

## Javascript

 ``

Output:

`20`

Time Complexity: O(n)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up