Replacing an element makes array elements consecutive

Given an array of positive distinct integers. We need to find the only element whose replacement with any other value makes array elements distinct consecutive. If it is not possible to make array elements consecutive, return -1.

Examples :

```Input : arr[] = {45, 42, 46, 48, 47}
Output : 42
Explanation: We can replace 42 with either
44 or 48 to make array consecutive.

Input : arr[] = {5, 6, 7, 9, 10}
Output : 5 [OR 10]
Explanation: We can either replace 5 with 8
or 10 with 8 to make array elements
consecutive.

Input : arr[] = {5, 6, 7, 9, 8}
Output : Array elements are already consecutive```

A Naive Approach is to check each element of arr[], after replacing of which makes consecutive or not. Time complexity for this approach O(n2)

A Better Approach is based on an important observation that either the smallest or the largest element would be answer if answer exists. If answer exists, then there are two cases.

1) Series of consecutive elements starts with minimum element of array then continues by adding 1 to previous.
2) Series of consecutive elements start with maximum element of array, then continues by subtracting 1 from previous.
We make above two series and for every series, we search series elements in array. If for both series, number of mismatches are more than 1, then answer does not exist. If any series is found with one mismatch, then we have answer.

C++

 `// C++ program to find an element replacement``// of which makes the array elements consecutive.``#include ``using` `namespace` `std;` `int` `findElement(``int` `arr[], ``int` `n)``{``    ``sort(arr, arr+n);`` ` `    ``// Making a series starting from first element``    ``// and adding 1 to every element.   ``    ``int` `mismatch_count1 = 0, res;``    ``int` `next_element = arr[n-1] - n + 1;``    ``for` `(``int` `i=0; i=1; i--) {``       ``if` `(binary_search(arr, arr+n, next_element) == 0)``       ``{``          ``res = arr[n-1]; ``          ``mismatch_count2++;``       ``}     ``       ``next_element--;``    ``}``        ` `    ``// If only one mismatch is found.``    ``if` `(mismatch_count2 == 1)``      ``return` `res;``      ` `    ``return` `-1; ``}` `// Driver code``int` `main()``{``    ``int` `arr[] =  {7, 5, 12, 8} ;``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``    ``int` `res = findElement(arr,n);``    ``if` `(res == -1)``      ``cout << ``"Answer does not exist"``;``    ``else` `if` `(res == 0)``      ``cout << ``"Elements are already consecutive"``;``    ``else``      ``cout << res;``    ``return` `0;``}`

Java

 `// Java program to find an element``// replacement of which makes``// the array elements consecutive.``import` `java.io.*;``import` `java.util.Arrays;` `class` `GFG``{``    ``static` `int` `findElement(``int` `[]arr,``                           ``int` `n)``    ``{``        ``Arrays.sort(arr);``    ` `        ``// Making a series starting``        ``// from first element and``        ``// adding 1 to every element.``        ``int` `mismatch_count1 = ``0``,``                        ``res = ``0``;``        ``int` `next_element = arr[n - ``1``] -``                               ``n + ``1``;``        ` `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++)``        ``{``        ``if` `(Arrays.binarySearch(arr,``                            ``next_element) < ``0``)``        ``{``            ``res = arr[``0``];``            ``mismatch_count1++;``        ``}``            ``next_element++;``        ``}``    ` `        ``// If only one mismatch is found.``        ``if` `(mismatch_count1 == ``1``)``            ``return` `res;``    ` `        ``// If no mismatch found, elements``        ``// are already consecutive.``        ``if` `(mismatch_count1 == ``0``)``            ``return` `0``;``    ` `        ``// Making a series starting``        ``// from last element and``        ``// subtracting 1 to every element.``        ``int` `mismatch_count2 = ``0``;``        ``next_element = arr[``0``] + n - ``1``;``        ` `        ``for` `(``int` `i = n - ``1``; i >= ``1``; i--)``        ``{``        ``if` `(Arrays.binarySearch(arr,``                            ``next_element) < ``0``)``        ``{``            ``res = arr[n - ``1``];``            ``mismatch_count2++;``        ``}    ``        ``next_element--;``        ``}``            ` `        ``// If only one mismatch is found.``        ``if` `(mismatch_count2 == ``1``)``        ``return` `res;``            ` `        ``return` `-``1``;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `[]arr = ``new` `int``[]{``7``, ``5``, ``12``, ``8``} ;``        ``int` `n = arr.length;``        ``int` `res = findElement(arr,n);``        ``if` `(res == -``1``)``        ``System.out.print(``"Answer does not exist"``);``        ``else` `if` `(res == ``0``)``        ``System.out.print(``"Elements are "` `+``                         ``"already consecutive"``);``        ``else``        ``System.out.print(res);``    ``}``}` `// This code is contributed by``// Manish Shaw(manishshaw1)`

Python3

 `# Python3 program to find an element``# replacement of which makes the``# array elements consecutive.``from` `bisect ``import` `bisect_left``  ` `def` `BinarySearch(a, x):``    ` `    ``i ``=` `bisect_left(a, x)``    ``if` `i !``=` `len``(a) ``and` `a[i] ``=``=` `x:``        ``return` `i``    ``else``:``        ``return` `-``1``  ` `def` `findElement(arr, n):``    ` `    ``arr.sort()``   ` `    ``# Making a series starting``    ``# from first element and``    ``# adding 1 to every element.    ``    ``mismatch_count1 ``=` `0``    ``res ``=` `0``    ``next_element ``=` `arr[n ``-` `1``] ``-` `n ``+` `1``    ` `    ``for` `i ``in` `range``(n ``-` `1``):``        ``if` `(BinarySearch(arr, next_element) ``=``=` `-``1``):``            ``res ``=` `arr[``0``]``            ``mismatch_count1 ``+``=` `1``            ` `        ``next_element ``+``=` `1` `    ``# If only one mismatch is found.``    ``if` `(mismatch_count1 ``=``=` `1``):``        ``return` `res``  ` `    ``# If no mismatch found, elements are``    ``# already consecutive.``    ``if` `(mismatch_count1 ``=``=` `0``):``        ``return` `0``  ` `    ``# Making a series starting from last element``    ``# and subtracting 1 to every element. ``    ``mismatch_count2 ``=` `0``    ``next_element ``=` `arr[``0``] ``+` `n ``-` `1``    ` `    ``for` `i ``in` `range``(n ``-` `1``, ``0``, ``-``1``):``        ``if` `BinarySearch(arr, next_element) ``=``=` `-``1``:``            ``res ``=` `arr[n ``-` `1``]  ``            ``mismatch_count2 ``+``=` `1``        ` `        ``next_element ``-``=` `1``    ` `    ``# If only one mismatch is found.``    ``if``(mismatch_count2 ``=``=` `1``):``      ``return` `res``        ` `    ``return` `-``1` `# Driver code``if` `__name__``=``=``"__main__"``:``    ` `    ``arr ``=`  `[ ``7``, ``5``, ``12``, ``8` `]``    ``n ``=` `len``(arr)``    ` `    ``res ``=` `findElement(arr, n)``    ` `    ``if` `(res ``=``=` `-``1``):``        ``print``(``"Answer does not exist"``)``    ``elif` `(res ``=``=` `0``):``        ``print``(``"Elements are already consecutive"``)``    ``else``:``        ``print``(res)``   ` `# This code is contributed by rutvik_56`

C#

 `// C# program to find an element``// replacement of which makes``// the array elements consecutive.``using` `System;``using` `System.Linq;``using` `System.Collections.Generic;` `class` `GFG``{``    ``static` `int` `findElement(``int` `[]arr,``                           ``int` `n)``    ``{``        ``Array.Sort(arr);``    ` `        ``// Making a series starting``        ``// from first element and``        ``// adding 1 to every element.``        ``int` `mismatch_count1 = 0, res = 0;``        ``int` `next_element = arr[n - 1] - n + 1;``        ` `        ``for` `(``int` `i = 0; i < n - 1; i++)``        ``{``        ``if` `(Array.BinarySearch(arr,``                               ``next_element) < 0)``        ``{``            ``res = arr[0];``            ``mismatch_count1++;``        ``}``            ``next_element++;``        ``}``    ` `        ``// If only one mismatch is found.``        ``if` `(mismatch_count1 == 1)``            ``return` `res;``    ` `        ``// If no mismatch found, elements``        ``// are already consecutive.``        ``if` `(mismatch_count1 == 0)``            ``return` `0;``    ` `        ``// Making a series starting``        ``// from last element and``        ``// subtracting 1 to every element.``        ``int` `mismatch_count2 = 0;``        ``next_element = arr[0] + n - 1;``        ` `        ``for` `(``int` `i = n - 1; i >= 1; i--)``        ``{``        ``if` `(Array.BinarySearch(arr,``                               ``next_element) < 0)``        ``{``            ``res = arr[n - 1];``            ``mismatch_count2++;``        ``}    ``        ``next_element--;``        ``}``            ` `        ``// If only one mismatch is found.``        ``if` `(mismatch_count2 == 1)``        ``return` `res;``            ` `        ``return` `-1;``    ``}``    ` `    ``// Driver code``    ``static` `void` `Main()``    ``{``        ``int` `[]arr = ``new` `int``[]{7, 5, 12, 8} ;``        ``int` `n = arr.Length;``        ``int` `res = findElement(arr,n);``        ``if` `(res == -1)``        ``Console.Write(``"Answer does not exist"``);``        ``else` `if` `(res == 0)``        ``Console.Write(``"Elements are "` `+``                      ``"already consecutive"``);``        ``else``        ``Console.Write(res);``    ``}``}` `// This code is contributed by``// Manish Shaw(manishshaw1)`

Javascript

 ``

Output :

`12`

Time Complexity: O(n Log n)
Auxiliary Space: O(1)

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