# Minimum increments to convert to an array of consecutive integers

Given an array arr[] with N elements, the task is to find the minimum number of operations required so that an Arithmetic Progression with the array elements is achieved with common difference as 1. In a single operation, any element can be incremented by 1.

Examples:

Input: arr[] = {4, 4, 5, 5, 7}
Output: 5
Desired array is {4, 5, 6, 7, 8} which
can be achieved in minimum possible operations.

Input: arr[] = {11, 2, 5, 6}
Output: 26
Since we are allowed to do only increment, we
change the array to {11, 12, 13, 14}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• We can utilize Binary Search to solve this problem.
• We will built the desired array with fixed last element which will be increased if the solution isn’t valid or decreased if its valid just like in binary search.
• Check that all the elements of desired array are greater than or equal to input array in order to perform operations on the elements to make them equal to the desired elements. Find the count of operations.
• Find the minimum possible value of last element which satisfy the condition of all the elements in the desired array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that return true if the ` `// required array can be generated ` `// with m as the last element ` `bool` `check(``int` `m, ``int` `n, ``int` `arr[]) ` `{ ` `    ``// Build the desired array ` `    ``int` `desired[n]; ` `    ``for` `(``int` `i = n - 1; i >= 0; i--) { ` `        ``desired[i] = m; ` `        ``m--; ` `    ``} ` ` `  `    ``// Check if the given array can ` `    ``// be converted to the desired array ` `    ``// with the given operation ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] > desired[i] || desired[i] < 1) { ` `            ``return` `false``; ` `        ``} ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// Function to return the minimum number ` `// of operations required to convert the ` `// given array to an increasing AP series ` `// with common difference as 1 ` `int` `minOperations(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `start = (``int``)arr[n - 1]; ` `    ``int` `end = *(max_element(arr, arr + n)) + n; ` `    ``int` `max_arr = 0; ` ` `  `    ``// Apply Binary Search ` `    ``while` `(start <= end) { ` `        ``int` `mid = (start + end) / 2; ` ` `  `        ``// If array can be generated with ` `        ``// mid as the last element ` `        ``if` `(check(mid, n, arr)) { ` ` `  `            ``// Current ans is mid ` `            ``max_arr = mid; ` ` `  `            ``// Check whether the same can be ` `            ``// achieved with even less operations ` `            ``end = mid - 1; ` `        ``} ` `        ``else` `{ ` `            ``start = mid + 1; ` `        ``} ` `    ``} ` ` `  `    ``// Build the desired array ` `    ``int` `desired[n]; ` `    ``for` `(``int` `i = n - 1; i >= 0; i--) { ` `        ``desired[i] = max_arr; ` `        ``max_arr--; ` `    ``} ` ` `  `    ``// Calculate the number of ` `    ``// operations required ` `    ``int` `operations = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``operations += (desired[i] - arr[i]); ` `    ``} ` ` `  `    ``// Return the number of ` `    ``// operations required ` `    ``return` `operations; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 4, 4, 5, 5, 7 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``cout << minOperations(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.Arrays; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function that return true if the ` `    ``// required array can be generated ` `    ``// with m as the last element ` `    ``static` `boolean` `check(``int` `m, ``int` `n, ``int` `arr[])  ` `    ``{ ` `        ``// Build the desired array ` `        ``int``[] desired = ``new` `int``[n]; ` `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--)  ` `        ``{ ` `            ``desired[i] = m; ` `            ``m--; ` `        ``} ` ` `  `        ``// Check if the given array can ` `        ``// be converted to the desired array ` `        ``// with the given operation ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``if` `(arr[i] > desired[i] || desired[i] < ``1``)  ` `            ``{ ` `                ``return` `false``; ` `            ``} ` `        ``} ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Function to return the minimum number ` `    ``// of operations required to convert the ` `    ``// given array to an increasing AP series ` `    ``// with common difference as 1 ` `    ``static` `int` `minOperations(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``int` `start = (``int``) arr[n - ``1``]; ` `        ``int` `end = Arrays.stream(arr).max().getAsInt() + n; ` `        ``int` `max_arr = ``0``; ` ` `  `        ``// Apply Binary Search ` `        ``while` `(start <= end)  ` `        ``{ ` `            ``int` `mid = (start + end) / ``2``; ` ` `  `            ``// If array can be generated with ` `            ``// mid as the last element ` `            ``if` `(check(mid, n, arr)) ` `            ``{ ` ` `  `                ``// Current ans is mid ` `                ``max_arr = mid; ` ` `  `                ``// Check whether the same can be ` `                ``// achieved with even less operations ` `                ``end = mid - ``1``; ` `            ``}  ` `            ``else` `            ``{ ` `                ``start = mid + ``1``; ` `            ``} ` `        ``} ` ` `  `        ``// Build the desired array ` `        ``int``[] desired = ``new` `int``[n]; ` `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--)  ` `        ``{ ` `            ``desired[i] = max_arr; ` `            ``max_arr--; ` `        ``} ` ` `  `        ``// Calculate the number of ` `        ``// operations required ` `        ``int` `operations = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` `            ``operations += (desired[i] - arr[i]); ` `        ``} ` ` `  `        ``// Return the number of ` `        ``// operations required ` `        ``return` `operations; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``4``, ``4``, ``5``, ``5``, ``7``}; ` `        ``int` `n = arr.length; ` ` `  `        ``System.out.println(minOperations(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function that return true if the  ` `# required array can be generated  ` `# with m as the last element  ` `def` `check( m, n, arr) : ` ` `  `    ``# Build the desired array  ` `    ``desired ``=` `[``0``]``*``n;  ` `    ``for` `i ``in` `range``(n``-``1``,``-``1``,``-``1``) : ` `        ``desired[i] ``=` `m;  ` `        ``m ``-``=` `1``;  ` `     `  ` `  `    ``# Check if the given array can  ` `    ``# be converted to the desired array  ` `    ``# with the given operation  ` `    ``for` `i ``in` `range``(n) :  ` `        ``if` `(arr[i] > desired[i] ``or` `desired[i] < ``1``) :  ` `            ``return` `False``;  ` ` `  `    ``return` `True` ` `  ` `  `# Function to return the minimum number  ` `# of operations required to convert the  ` `# given array to an increasing AP series  ` `# with common difference as 1  ` `def` `minOperations(arr, n) :  ` ` `  `    ``start ``=` `arr[n ``-` `1``];  ` `    ``end ``=` `max``(arr) ``+` `n;  ` `    ``max_arr ``=` `0``;  ` ` `  `    ``# Apply Binary Search  ` `    ``while` `(start <``=` `end) :  ` `        ``mid ``=` `(start ``+` `end) ``/``/` `2``;  ` ` `  `        ``# If array can be generated with  ` `        ``# mid as the last element  ` `        ``if` `(check(mid, n, arr)) : ` ` `  `            ``# Current ans is mid  ` `            ``max_arr ``=` `mid;  ` ` `  `            ``# Check whether the same can be  ` `            ``# achieved with even less operations  ` `            ``end ``=` `mid ``-` `1``;  ` `         `  `        ``else` `: ` `            ``start ``=` `mid ``+` `1``;  ` ` `  `    ``# Build the desired array  ` `    ``desired ``=` `[``0``]``*` `n;  ` `    ``for` `i ``in` `range``(n``-``1``, ``-``1``,``-``1``) : ` `        ``desired[i] ``=` `max_arr;  ` `        ``max_arr ``-``=` `1``;  ` `     `  ` `  `    ``# Calculate the number of  ` `    ``# operations required  ` `    ``operations ``=` `0``;  ` `    ``for` `i ``in` `range``(n) : ` `        ``operations ``+``=` `(desired[i] ``-` `arr[i]);  ` `     `  `    ``# Return the number of  ` `    ``# operations required  ` `    ``return` `operations;  ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``4``, ``4``, ``5``, ``5``, ``7` `];  ` `    ``n ``=` `len``(arr);  ` ` `  `    ``print``(minOperations(arr, n));  ` `     `  `    ``# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System;      ` `using` `System.Linq;  ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function that return true if the ` `    ``// required array can be generated ` `    ``// with m as the last element ` `    ``static` `Boolean check(``int` `m, ``int` `n, ``int` `[]arr)  ` `    ``{ ` `        ``// Build the desired array ` `        ``int``[] desired = ``new` `int``[n]; ` `        ``for` `(``int` `i = n - 1; i >= 0; i--)  ` `        ``{ ` `            ``desired[i] = m; ` `            ``m--; ` `        ``} ` ` `  `        ``// Check if the given array can ` `        ``// be converted to the desired array ` `        ``// with the given operation ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `(arr[i] > desired[i] || desired[i] < 1)  ` `            ``{ ` `                ``return` `false``; ` `            ``} ` `        ``} ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Function to return the minimum number ` `    ``// of operations required to convert the ` `    ``// given array to an increasing AP series ` `    ``// with common difference as 1 ` `    ``static` `int` `minOperations(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``int` `start = (``int``) arr[n - 1]; ` `        ``int` `end = arr.Max() + n; ` `        ``int` `max_arr = 0; ` ` `  `        ``// Apply Binary Search ` `        ``while` `(start <= end)  ` `        ``{ ` `            ``int` `mid = (start + end) / 2; ` ` `  `            ``// If array can be generated with ` `            ``// mid as the last element ` `            ``if` `(check(mid, n, arr)) ` `            ``{ ` ` `  `                ``// Current ans is mid ` `                ``max_arr = mid; ` ` `  `                ``// Check whether the same can be ` `                ``// achieved with even less operations ` `                ``end = mid - 1; ` `            ``}  ` `            ``else` `            ``{ ` `                ``start = mid + 1; ` `            ``} ` `        ``} ` ` `  `        ``// Build the desired array ` `        ``int``[] desired = ``new` `int``[n]; ` `        ``for` `(``int` `i = n - 1; i >= 0; i--)  ` `        ``{ ` `            ``desired[i] = max_arr; ` `            ``max_arr--; ` `        ``} ` ` `  `        ``// Calculate the number of ` `        ``// operations required ` `        ``int` `operations = 0; ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``operations += (desired[i] - arr[i]); ` `        ``} ` ` `  `        ``// Return the number of ` `        ``// operations required ` `        ``return` `operations; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``int` `[]arr = {4, 4, 5, 5, 7}; ` `        ``int` `n = arr.Length; ` ` `  `        ``Console.WriteLine(minOperations(arr, n)); ` `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

Output:

```5
```

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