# Maximizing the elements with a[i+1] > a[i]

Given an array of N integers, rearrange the array elements such that the next array element is greater than the previous element ( > ).

Examples:

Input : arr[] = {20, 30, 10, 50, 40}
Output : 4
We rearrange the array as 10, 20, 30, 40, 50. As 20 > 10, 30 > 20, 40 > 30, 50 > 40, so we get 4 indices i such that > .

Input : arr[] = {200, 100, 100, 200}
Output : 2
We get optimal arrangement as 100 200 100 200.

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

If all elements are distinct, then answer is simply n-1 where n is the number of elements in the array. If there are repeating elements, then answer is n – max_freq.

## Java

 `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// returns the number of positions where A(i + 1) is ` `    ``// greater than A(i) after rearrangement of the array ` `    ``static` `int` `countMaxPos(``int``[] arr) ` `    ``{ ` `        ``int` `n = arr.length; ` ` `  `        ``// Creating a HashMap containing char ` `        ``// as a key and occurrences as  a value ` `        ``HashMap map ` `            ``= ``new` `HashMap(); ` `        ``for` `(``int` `x : arr) { ` `            ``if` `(map.containsKey(x)) ` `                ``map.put(x, map.get(x) + ``1``); ` `            ``else` `                ``map.put(x, ``1``); ` `        ``} ` ` `  `        ``// Find the maximum frequency ` `        ``int` `max_freq = ``0``; ` `        ``for` `(Map.Entry entry : map.entrySet()) ` `            ``max_freq = Math.max(max_freq, (``int``)entry.getValue()); ` ` `  `        ``return` `n - max_freq; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] arr = { ``20``, ``30``, ``10``, ``50``, ``40` `}; ` `        ``System.out.println(countMaxPos(arr)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the above approach ` ` `  `# Returns the number of positions where  ` `# A(i + 1) is greater than A(i) after  ` `# rearrangement of the array  ` `def` `countMaxPos(arr):  ` `     `  `    ``n ``=` `len``(arr)  ` ` `  `    ``# Creating a HashMap containing char  ` `    ``# as a key and occurrences as a value  ` `    ``Map` `=` `{}  ` `    ``for` `x ``in` `arr:  ` `        ``if` `x ``in` `Map``:  ` `            ``Map``[x] ``+``=` `1` `        ``else``: ` `            ``Map``[x] ``=` `1` `         `  `    ``# Find the maximum frequency  ` `    ``max_freq ``=` `0` `    ``for` `entry ``in` `Map``:  ` `        ``max_freq ``=` `max``(max_freq, ``Map``[entry])  ` ` `  `    ``return` `n ``-` `max_freq  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``arr ``=` `[``20``, ``30``, ``10``, ``50``, ``40``]  ` `    ``print``(countMaxPos(arr))  ` `     `  `# This code is contributed by Rituraj Jain `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic;              ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// returns the number of positions where  ` `    ``// A(i + 1) is greater than A(i) after  ` `    ``// rearrangement of the array ` `    ``static` `int` `countMaxPos(``int``[] arr) ` `    ``{ ` `        ``int` `n = arr.Length; ` ` `  `        ``// Creating a HashMap containing char ` `        ``// as a key and occurrences as a value ` `        ``Dictionary<``int``,  ` `                   ``int``> map = ``new` `Dictionary<``int``, ` `                                             ``int``>(); ` `        ``foreach` `(``int` `x ``in` `arr)  ` `        ``{ ` `            ``if` `(map.ContainsKey(x)) ` `                ``map[x] = map[x] + 1; ` `            ``else` `                ``map.Add(x, 1); ` `        ``} ` ` `  `        ``// Find the maximum frequency ` `        ``int` `max_freq = 0; ` `        ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `map) ` `            ``max_freq = Math.Max(max_freq, entry.Value); ` ` `  `        ``return` `n - max_freq; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int``[] arr = { 20, 30, 10, 50, 40 }; ` `        ``Console.WriteLine(countMaxPos(arr)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```4
```

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Improved By : rituraj_jain, 29AjayKumar

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