# Minimum index i such that all the elements from index i to given index are equal

Given an array arr[] of integers and an integer pos, the task is to find the minimum index i such that all the elements from index i to index pos are equal.

Examples:

Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 3
Output:
Elements in index range [1, 3] are all equal to 1.
Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 5
Output:

Simple Approach: Starting from index pos – 1, traverse the array in reverse and for the first index i such that arr[i] != arr[pos] print i + 1 which is the required index.
Below is the implementation of the above approach:

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the minimum required index` `int` `minIndex(``int` `arr[], ``int` `n, ``int` `pos)` `{` `    ``int` `num = arr[pos];`   `    ``// Start from arr[pos - 1]` `    ``int` `i = pos - 1;` `    ``while` `(i >= 0) {` `        ``if` `(arr[i] != num)` `            ``break``;` `        ``i--;` `    ``}`   `    ``// All elements are equal` `    ``// from arr[i + 1] to arr[pos]` `    ``return` `i + 1;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 2, 1, 1, 1, 5, 2 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``int` `pos = 4;` `    `  `      ``// Function Call` `    ``cout << minIndex(arr, n, pos);` `    ``return` `0;` `}`

 `// Java implementation of the approach` `class` `GFG {`   `    ``// Function to return the minimum required index` `    ``static` `int` `minIndex(``int` `arr[], ``int` `n, ``int` `pos)` `    ``{` `        ``int` `num = arr[pos];`   `        ``// Start from arr[pos - 1]` `        ``int` `i = pos - ``1``;` `        ``while` `(i >= ``0``) {` `            ``if` `(arr[i] != num)` `                ``break``;` `            ``i--;` `        ``}`   `        ``// All elements are equal` `        ``// from arr[i + 1] to arr[pos]` `        ``return` `i + ``1``;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``2``, ``1``, ``1``, ``1``, ``5``, ``2` `};` `        ``int` `n = arr.length;` `        ``int` `pos = ``4``;` `        `  `          ``// Function Call` `        ``System.out.println(minIndex(arr, n, pos));` `    ``}` `}`   `// This code is contributed by Code_Mech.`

 `# Python3 implementation of the approach`   `# Function to return the minimum` `# required index` `def` `minIndex(arr, n, pos):`   `    ``num ``=` `arr[pos]`   `    ``# Start from arr[pos - 1]` `    ``i ``=` `pos ``-` `1` `    ``while` `(i >``=` `0``):` `        ``if` `(arr[i] !``=` `num):` `            ``break` `        ``i ``-``=` `1` `    `  `    ``# All elements are equal` `    ``# from arr[i + 1] to arr[pos]` `    ``return` `i ``+` `1`   `# Driver code` `arr ``=` `[``2``, ``1``, ``1``, ``1``, ``5``, ``2` `]` `n ``=` `len``(arr)` `pos ``=` `4`   `# Function Call` `print``(minIndex(arr, n, pos))`   `# This code is contributed by ` `# Mohit Kumar 29`

 `// C# implementation of the approach` `using` `System;` `class` `GFG {`   `    ``// Function to return the minimum required index` `    ``static` `int` `minIndex(``int``[] arr, ``int` `n, ``int` `pos)` `    ``{` `        ``int` `num = arr[pos];`   `        ``// Start from arr[pos - 1]` `        ``int` `i = pos - 1;` `        ``while` `(i >= 0) {` `            ``if` `(arr[i] != num)` `                ``break``;` `            ``i--;` `        ``}`   `        ``// All elements are equal` `        ``// from arr[i + 1] to arr[pos]` `        ``return` `i + 1;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 2, 1, 1, 1, 5, 2 };` `        ``int` `n = arr.Length;` `        ``int` `pos = 4;` `        `  `          ``// Function Call` `        ``Console.WriteLine(minIndex(arr, n, pos));` `    ``}` `}`   `// This code is contributed` `// by Akanksha Rai`

 `= 0) ` `    ``{ ` `        ``if` `(``\$arr``[``\$i``] != ``\$num``) ` `            ``break``; ` `        ``\$i``--; ` `    ``} `   `    ``// All elements are equal ` `    ``// from arr[i + 1] to arr[pos] ` `    ``return` `\$i` `+ 1; ` `} `   `// Driver code ` `\$arr` `= ``array``(2, 1, 1, 1, 5, 2 ); ` `\$n` `= sizeof(``\$arr``); ` `\$pos` `= 4; `   `echo` `minIndex(``\$arr``, ``\$n``, ``\$pos``); `   `// This code is contributed by Ryuga` `?>`

Output

`4`

Time Complexity: O(N)
Space Complexity: O(1)

Efficient Approach :

Do a binary search in the sub-array [0, pos-1]. Stop condition will be if arr[mid] == arr[pos] && arr[mid-1] != arr[pos]. Go-left or Go-right will depend on if arr[mid] == arr[pos] or not respectively.

Implementation:

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the minimum required index` `int` `minIndex(``int` `arr[], ``int` `pos)` `{` `    ``int` `low = 0;` `    ``int` `high = pos;` `    ``int` `i = pos;`   `    ``while` `(low < high) {` `        ``int` `mid = (low + high) / 2;` `        ``if` `(arr[mid] != arr[pos]) {` `            ``low = mid + 1;` `        ``}` `        ``else` `{` `            ``high = mid - 1;` `            ``i = mid;` `            ``if` `(mid > 0 && arr[mid - 1] != arr[pos]) {`   `                ``// Short-cicuit more comparisions as found` `                ``// the border point` `                ``break``;` `            ``}` `        ``}` `    ``}`   `    ``// For cases were high = low + 1 and arr[high] will` `    ``// match with` `    ``// arr[pos] but not arr[low] or arr[mid]. In such` `    ``// iteration the if condition will satisfy and loop will` `    ``// break post that low will be updated. Hence i will not` `    ``// point to the correct index.` `    ``return` `arr[low] == arr[pos] ? low : i;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 2, 1, 1, 1, 5, 2 };`   `    ``cout << minIndex(arr, 2) << endl; ``// Should be 1` `    ``cout << minIndex(arr, 3) << endl; ``// Should be 1` `    ``cout << minIndex(arr, 4) << endl; ``// Should be 4` `    ``return` `0;` `}`   `// This code is contributed by` `// anshbikram`

 `# Python3 implementation of the approach`   `# Function to return the minimum` `# required index`   `def` `minIndex(arr, pos):` `    ``low ``=` `0` `    ``high ``=` `pos` `    ``i ``=` `pos`   `    ``while` `low < high:` `        ``mid ``=` `(low ``+` `high)``/``/``2` `        ``if` `arr[mid] !``=` `arr[pos]:` `            ``low ``=` `mid ``+` `1` `        ``else``:` `            ``high ``=` `mid ``-` `1` `            ``i ``=` `mid` `            ``if` `mid > ``0` `and` `arr[mid``-``1``] !``=` `arr[pos]:`   `                ``# Short-cicuit more comparisions as found the border point` `                ``break`   `    ``# For cases were high = low + 1 and arr[high] will match with` `    ``# arr[pos] but not arr[low] or arr[mid]. In such iteration` `    ``# the if condition will satisfy and loop will break post that` `    ``# low will be updated. Hence i will not point to the correct index.` `    ``return` `low ``if` `arr[low] ``=``=` `arr[pos] ``else` `i`     `# Driver code` `arr ``=` `[``2``, ``1``, ``1``, ``1``, ``5``, ``2``]`   `print``(minIndex(arr, ``2``))  ``# Should be 1` `print``(minIndex(arr, ``3``))  ``# Should be 1` `print``(minIndex(arr, ``4``))  ``# Should be 4`   `# This code is contributed by` `# anshbikram`

 `// Java implementation of the approach`   `class` `GFG {` `    `  `      ``// Function to return the minimum required index` `    ``static` `int` `minIndex(``int` `arr[], ``int` `pos)` `    ``{` `        ``int` `low = ``0``;` `        ``int` `high = pos;` `        ``int` `i = pos;`   `        ``while` `(low < high) {` `            ``int` `mid = (low + high) / ``2``;` `            ``if` `(arr[mid] != arr[pos]) {` `                ``low = mid + ``1``;` `            ``}` `            ``else` `{` `                ``high = mid - ``1``;` `                ``i = mid;` `                ``if` `(mid > ``0` `&& arr[mid - ``1``] != arr[pos]) {` `                    `  `                      ``// Short-cicuit more comparisions as` `                    ``// found the border point` `                    ``break``;` `                ``}` `            ``}` `        ``}`   `        ``// For cases were high = low + 1 and arr[high] will` `        ``// match with arr[pos] but not arr[low] or arr[mid].` `        ``// In such iteration the if condition will satisfy` `        ``// and loop will break post that low will be` `        ``// updated. Hence i will not point to the correct` `        ``// index.` `        ``return` `arr[low] == arr[pos] ? low : i;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``2``, ``1``, ``1``, ``1``, ``5``, ``2` `};`   `        ``System.out.println(minIndex(arr, ``2``)); ``// Should be 1` `        ``System.out.println(minIndex(arr, ``3``)); ``// Should be 1` `        ``System.out.println(minIndex(arr, ``4``)); ``// Should be 4` `    ``}` `}`   `// This code is contributed by` `// anshbikram`

Output
```1
1
4
```

Time Complexity: O(log(n))
Space Complexity: O(1)

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