# Maximum distance between two unequal elements

Given an array **arr[]**, the task is to find the maximum distance between two unequal elements of the given array.**Examples:**

Input:arr[] = {3, 2, 1, 2, 1}Output:4

The maximum distance is between the first and the last element.Input:arr[] = {3, 3, 1, 3, 3}Output:2

**Naive approach:** Traverse the whole array for every single element and find the longest distance of element which is unequal.**Efficient approach:** By using the fact that the pair of unequal elements must include either first or last element or both, calculate the longest distance between unequal element by traversing the array either by fixing the first element or fixing the last element.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the maximum distance` `// between two unequal elements` `int` `maxDistance(` `int` `arr[], ` `int` `n)` `{` ` ` `// If first and last elements are unequal` ` ` `// they are maximum distance apart` ` ` `if` `(arr[0] != arr[n - 1])` ` ` `return` `(n - 1);` ` ` `int` `i = n - 1;` ` ` `// Fix first element as one of the elements` ` ` `// and start traversing from the right` ` ` `while` `(i > 0) {` ` ` `// Break for the first unequal element` ` ` `if` `(arr[i] != arr[0])` ` ` `break` `;` ` ` `i--;` ` ` `}` ` ` `// To store the distance from the first element` ` ` `int` `distFirst = (i == 0) ? -1 : i;` ` ` `i = 0;` ` ` `// Fix last element as one of the elements` ` ` `// and start traversing from the left` ` ` `while` `(i < n - 1) {` ` ` `// Break for the first unequal element` ` ` `if` `(arr[i] != arr[n - 1])` ` ` `break` `;` ` ` `i++;` ` ` `}` ` ` `// To store the distance from the last element` ` ` `int` `distLast = (i == n - 1) ? -1 : (n - 1 - i);` ` ` `// Maximum possible distance` ` ` `int` `maxDist = max(distFirst, distLast);` ` ` `return` `maxDist;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 4, 4, 1, 2, 1, 4 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << maxDistance(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.io.*;` `class` `GFG` `{` `// Function to return the maximum distance` `// between two unequal elements` `static` `int` `maxDistance(` `int` `arr[], ` `int` `n)` `{` ` ` `// If first and last elements are unequal` ` ` `// they are maximum distance apart` ` ` `if` `(arr[` `0` `] != arr[n - ` `1` `])` ` ` `return` `(n - ` `1` `);` ` ` `int` `i = n - ` `1` `;` ` ` `// Fix first element as one of the elements` ` ` `// and start traversing from the right` ` ` `while` `(i > ` `0` `)` ` ` `{` ` ` `// Break for the first unequal element` ` ` `if` `(arr[i] != arr[` `0` `])` ` ` `break` `;` ` ` `i--;` ` ` `}` ` ` `// To store the distance from the first element` ` ` `int` `distFirst = (i == ` `0` `) ? -` `1` `: i;` ` ` `i = ` `0` `;` ` ` `// Fix last element as one of the elements` ` ` `// and start traversing from the left` ` ` `while` `(i < n - ` `1` `)` ` ` `{` ` ` `// Break for the first unequal element` ` ` `if` `(arr[i] != arr[n - ` `1` `])` ` ` `break` `;` ` ` `i++;` ` ` `}` ` ` `// To store the distance from the last element` ` ` `int` `distLast = (i == n - ` `1` `) ? -` `1` `: (n - ` `1` `- i);` ` ` `// Maximum possible distance` ` ` `int` `maxDist = Math.max(distFirst, distLast);` ` ` `return` `maxDist;` `}` `// Driver code` `public` `static` `void` `main (String[] args)` `{` ` ` `int` `arr[] = { ` `4` `, ` `4` `, ` `1` `, ` `2` `, ` `1` `, ` `4` `};` ` ` `int` `n = arr.length;` ` ` `System.out.print(maxDistance(arr, n));` `}` `}` `// This code is contributed by anuj_67..` |

## Python3

`# Python implementation of the approach` `# Function to return the maximum distance` `# between two unequal elements` `def` `maxDistance(arr, n):` ` ` ` ` `# If first and last elements are unequal` ` ` `# they are maximum distance apart` ` ` `if` `(arr[` `0` `] !` `=` `arr[n ` `-` `1` `]):` ` ` `return` `(n ` `-` `1` `);` ` ` `i ` `=` `n ` `-` `1` `;` ` ` `# Fix first element as one of the elements` ` ` `# and start traversing from the right` ` ` `while` `(i > ` `0` `):` ` ` `# Break for the first unequal element` ` ` `if` `(arr[i] !` `=` `arr[` `0` `]):` ` ` `break` `;` ` ` `i` `-` `=` `1` `;` ` ` `# To store the distance from the first element` ` ` `distFirst ` `=` `-` `1` `if` `(i ` `=` `=` `0` `) ` `else` `i;` ` ` `i ` `=` `0` `;` ` ` `# Fix last element as one of the elements` ` ` `# and start traversing from the left` ` ` `while` `(i < n ` `-` `1` `):` ` ` `# Break for the first unequal element` ` ` `if` `(arr[i] !` `=` `arr[n ` `-` `1` `]):` ` ` `break` `;` ` ` `i` `+` `=` `1` `;` ` ` `# To store the distance from the last element` ` ` `distLast ` `=` `-` `1` `if` `(i ` `=` `=` `n ` `-` `1` `) ` `else` `(n ` `-` `1` `-` `i);` ` ` `# Maximum possible distance` ` ` `maxDist ` `=` `max` `(distFirst, distLast);` ` ` `return` `maxDist;` `# Driver code` `arr ` `=` `[` `4` `, ` `4` `, ` `1` `, ` `2` `, ` `1` `, ` `4` `];` `n ` `=` `len` `(arr);` `print` `(maxDistance(arr, n));` `# This code has been contributed by 29AjayKumar` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to return the maximum distance` `// between two unequal elements` `static` `int` `maxDistance(` `int` `[]arr, ` `int` `n)` `{` ` ` `// If first and last elements are unequal` ` ` `// they are maximum distance apart` ` ` `if` `(arr[0] != arr[n - 1])` ` ` `return` `(n - 1);` ` ` `int` `i = n - 1;` ` ` `// Fix first element as one of the elements` ` ` `// and start traversing from the right` ` ` `while` `(i > 0)` ` ` `{` ` ` `// Break for the first unequal element` ` ` `if` `(arr[i] != arr[0])` ` ` `break` `;` ` ` `i--;` ` ` `}` ` ` `// To store the distance from the first element` ` ` `int` `distFirst = (i == 0) ? -1 : i;` ` ` `i = 0;` ` ` `// Fix last element as one of the elements` ` ` `// and start traversing from the left` ` ` `while` `(i < n - 1)` ` ` `{` ` ` `// Break for the first unequal element` ` ` `if` `(arr[i] != arr[n - 1])` ` ` `break` `;` ` ` `i++;` ` ` `}` ` ` `// To store the distance from the last element` ` ` `int` `distLast = (i == n - 1) ? -1 : (n - 1 - i);` ` ` `// Maximum possible distance` ` ` `int` `maxDist = Math.Max(distFirst, distLast);` ` ` `return` `maxDist;` `}` `// Driver code` `static` `public` `void` `Main ()` `{` ` ` `int` `[]arr = { 4, 4, 1, 2, 1, 4 };` ` ` `int` `n = arr.Length;` ` ` `Console.WriteLine(maxDistance(arr, n));` `}` `}` `// This code is contributed by Tushil..` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to return the maximum distance` `// between two unequal elements` `function` `maxDistance(arr, n)` `{` ` ` `// If first and last elements are unequal` ` ` `// they are maximum distance apart` ` ` `if` `(arr[0] != arr[n - 1])` ` ` `return` `(n - 1);` ` ` `var` `i = n - 1;` ` ` `// Fix first element as one of the elements` ` ` `// and start traversing from the right` ` ` `while` `(i > 0) {` ` ` `// Break for the first unequal element` ` ` `if` `(arr[i] != arr[0])` ` ` `break` `;` ` ` `i--;` ` ` `}` ` ` `// To store the distance from the first element` ` ` `var` `distFirst = (i == 0) ? -1 : i;` ` ` `i = 0;` ` ` `// Fix last element as one of the elements` ` ` `// and start traversing from the left` ` ` `while` `(i < n - 1) {` ` ` `// Break for the first unequal element` ` ` `if` `(arr[i] != arr[n - 1])` ` ` `break` `;` ` ` `i++;` ` ` `}` ` ` `// To store the distance from the last element` ` ` `var` `distLast = (i == n - 1) ? -1 : (n - 1 - i);` ` ` `// Maximum possible distance` ` ` `var` `maxDist = Math.max(distFirst, distLast);` ` ` `return` `maxDist;` `}` `// Driver code` `var` `arr = [ 4, 4, 1, 2, 1, 4 ];` `var` `n = arr.length;` `document.write( maxDistance(arr, n));` `</script>` |

**Output:**

4

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