# Minimum elements to be removed such that sum of adjacent elements is always even

Given an array of N integers. The task is to eliminate the minimum number of elements such that in the resulting array the sum of any two adjacent values is even.

**Examples:**

Input :arr[] = {1, 2, 3}Output :1 Remove 2 from the array.Input :arr[] = {1, 3, 5, 4, 2}Output :2 Remove 4 and 2.

**Approach:** The sum of 2 numbers is even if either both of them is odd or both of them is even. This means for every pair of consecutive numbers that have the different parity, eliminate one of them.

So, to make the adjacent elements sum even, either all elements should be odd or even. So the following greedy algorithm works:

- Go through all the elements in order.
- Count the odd and even elements in the array.
- Return the minimum count.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find minimum number of eliminations` `// such that sum of all adjacent elements is even` `int` `min_elimination(` `int` `n, ` `int` `arr[])` `{` ` ` `int` `countOdd = 0;` ` ` `// Stores the new value` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `// Count odd numbers` ` ` `if` `(arr[i] % 2)` ` ` `countOdd++;` ` ` `// Return the minimum of even and` ` ` `// odd count` ` ` `return` `min(countOdd, n - countOdd);` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 1, 2, 3, 7, 9 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << min_elimination(n, arr);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the above approach` `class` `GFG` `{` ` ` `// Function to find minimum number of ` `// eliminations such that sum of all ` `// adjacent elements is even` `static` `int` `min_elimination(` `int` `n, ` `int` `arr[])` `{` ` ` `int` `countOdd = ` `0` `;` ` ` `// Stores the new value` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `// Count odd numbers` ` ` `if` `(arr[i] % ` `2` `== ` `1` `)` ` ` `countOdd++;` ` ` `// Return the minimum of even ` ` ` `// and odd count` ` ` `return` `Math.min(countOdd, n - countOdd);` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[] = { ` `1` `, ` `2` `, ` `3` `, ` `7` `, ` `9` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(min_elimination(n, arr));` `}` `}` `// This code is contributed by Code_Mech` |

## Python3

`# Python 3 implementation of the ` `# above approach` `# Function to find minimum number of ` `# eliminations such that sum of all ` `# adjacent elements is even` `def` `min_elimination(n, arr):` ` ` `countOdd ` `=` `0` ` ` `# Stores the new value` ` ` `for` `i ` `in` `range` `(n):` ` ` ` ` `# Count odd numbers` ` ` `if` `(arr[i] ` `%` `2` `):` ` ` `countOdd ` `+` `=` `1` ` ` `# Return the minimum of even and` ` ` `# odd count` ` ` `return` `min` `(countOdd, n ` `-` `countOdd)` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `arr ` `=` `[` `1` `, ` `2` `, ` `3` `, ` `7` `, ` `9` `]` ` ` `n ` `=` `len` `(arr)` ` ` `print` `(min_elimination(n, arr))` `# This code is contributed by` `# Surendra_Gangwar` |

## C#

`// C# implementation of the above approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to find minimum number of ` `// eliminations such that sum of all ` `// adjacent elements is even` `static` `int` `min_elimination(` `int` `n, ` `int` `[] arr)` `{` ` ` `int` `countOdd = 0;` ` ` `// Stores the new value` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `// Count odd numbers` ` ` `if` `(arr[i] % 2 == 1)` ` ` `countOdd++;` ` ` `// Return the minimum of even ` ` ` `// and odd count` ` ` `return` `Math.Min(countOdd, n - countOdd);` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `[] arr = { 1, 2, 3, 7, 9 };` ` ` `int` `n = arr.Length;` ` ` `Console.WriteLine(min_elimination(n, arr));` `}` `}` `// This code is contributed by Code_Mech` |

## PHP

`<?php` `// PHP implementation of the above approach` `// Function to find minimum number of ` `// eliminations such that sum of all ` `// adjacent elements is even` `function` `min_elimination(` `$n` `, ` `$arr` `)` `{` ` ` `$countOdd` `= 0;` ` ` `// Stores the new value` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `// Count odd numbers` ` ` `if` `(` `$arr` `[` `$i` `] % 2 == 1)` ` ` `$countOdd` `++;` ` ` `// Return the minimum of even ` ` ` `// and odd count` ` ` `return` `min(` `$countOdd` `, ` `$n` `- ` `$countOdd` `);` `}` `// Driver code` `$arr` `= ` `array` `(1, 2, 3, 7, 9);` `$n` `= sizeof(` `$arr` `);` `echo` `(min_elimination(` `$n` `, ` `$arr` `));` `// This code is contributed by Code_Mech` `?>` |

## Javascript

`<script>` `// Function to find minimum number of eliminations` `// such that sum of all adjacent elements is even` `function` `min_elimination(n, arr)` `{` ` ` `let countOdd = 0;` ` ` `// Stores the new value` ` ` `for` `(let i = 0; i < n; i++)` ` ` `// Count odd numbers` ` ` `if` `(arr[i] % 2)` ` ` `countOdd++;` ` ` `// Return the minimum of even and` ` ` `// odd count` ` ` `return` `Math.min(countOdd, n - countOdd);` `}` `// Driver code` `let arr= [1, 2, 3, 7, 9];` `let n = arr.length;` `document.write(min_elimination(n, arr));` `</script>` |

**Output:**

1

**Time Complexity: **O(N )

**Auxiliary Space:** O(1)

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