# Minimum no. of operations required to make all Array Elements Zero

Given an array of N elements and each element is either 1 or 0. You need to make all the elements of the array equal to 0 by performing below operations:

• If an element is 1, You can change it’s value equal to 0 then,
• if the next consecutive element is 1, it will automatically get converted to 0.
• if the next consecutive element is already 0, nothing will happen.

Now, the task is to find the minimum number of operations required to make all elements equal to 0.

Examples:

```Input : arr[] = {1, 1, 0, 0, 1, 1, 1, 0, 0, 1}
Output : Minimum changes: 3

Input : arr[] = {1, 1, 1, 1}
Output : Minimum changes: 1
```

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

Approach: To find the minimum number of changes required, iterate the array from left to right and check if the current element is 1 or not. If the current element is 1, then change it to 0 and increment the count by 1 and search for the 0 for the next operation as all consecutive 1’s will be automatically converted to 0.

Below is the implementation of above approach:

## C++

 `// CPP program to find  minimum number of  ` `// operations required to make all  ` `// array elements zero ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find  minimum number of  ` `// operations required to make all  ` `// array elements zero ` `int` `minimumChanges(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `i; ` `     `  `    ``// It will maintain total changes required ` `    ``int` `changes = 0;  ` `     `  `    ``for` `(i = 0; i < n; i++)  ` `    ``{    ` `        ``// Check for the first 1 ` `        ``if` `(arr[i] == 1)  ` `        ``{    ` `            ``int` `j; ` `             `  `            ``// Check for number of  ` `            ``// consecutive 1's ` `            ``for``(j = i+1; j

## Java

 `// Java program to find minimum  ` `// number of operations required  ` `// to make all array elements zero  ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to find minimum number  ` `// of operations required to make   ` `// all array elements zero  ` `static` `int` `minimumChanges(``int` `arr[],  ` `                          ``int` `n)  ` `{  ` `    ``int` `i;  ` `     `  `    ``// It will maintain total ` `    ``// changes required  ` `    ``int` `changes = ``0``;  ` `     `  `    ``for` `(i = ``0``; i < n; i++)  ` `    ``{  ` `        ``// Check for the first 1  ` `        ``if` `(arr[i] == ``1``)  ` `        ``{  ` `            ``int` `j;  ` `             `  `            ``// Check for number of  ` `            ``// consecutive 1's  ` `            ``for``(j = i + ``1``; j < n; j++)  ` `            ``{  ` `                ``if``(arr[j] == ``0``)  ` `                    ``break``;  ` `            ``}  ` `             `  `            ``// Increment i to the position   ` `            ``// of last consecutive 1  ` `            ``i = j - ``1``;  ` `             `  `            ``changes++;  ` `        ``}  ` `    ``}  ` `     `  `    ``return` `changes;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main (String args[]) ` `{ ` `    ``int` `arr[] = { ``1``, ``1``, ``0``, ``0``, ``0``,  ` `                     ``1``, ``0``, ``1``, ``1` `};  ` `    ``int` `n = arr.length ; ` `     `  `    ``System.out.println(``"Minimum operations: "` `+  ` `                        ``minimumChanges(arr, n));  ` `     `  `} ` `} ` ` `  `// This code is contributed by ANKITRAI1 `

## Python 3

 `# Python 3 program to find  ` `# minimum number of operations  ` `# required to make all array ` `# elements zero  ` ` `  `# Function to find minimum number  ` `# of operations required to make  ` `# all array elements zero  ` `def` `minimumChanges(arr, n) : ` ` `  `    ``# It will maintain total ` `    ``# changes required  ` `    ``changes ``=` `0` `     `  `    ``i ``=` `0` `     `  `    ``while` `i < n : ` ` `  `        ``# Check for the first 1 ` `        ``if` `arr[i] ``=``=` `1` `: ` ` `  `            ``j ``=` `i ``+` `1` ` `  `            ``# Check for number of  ` `            ``# consecutive 1's  ` `            ``while` `j < n: ` ` `  `                ``if` `arr[j] ``=``=` `0` `: ` `                    ``break` ` `  `                ``j ``+``=` `1` ` `  `            ``# Increment i to the position  ` `            ``# of last consecutive 1  ` `            ``i ``=` `j ``-` `1` `             `  `            ``changes ``+``=` `1` ` `  `        ``i ``+``=` `1` `         `  `    ``return` `changes ` ` `  `# Driver code      ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``arr ``=` `[ ``1``, ``1``, ``0``, ``0``, ``0``, ``1``, ``0``, ``1``, ``1``] ` `    ``n ``=` `len``(arr) ` ` `  `    ``print``(``"Minimum operations:"``,  ` `           ``minimumChanges(arr, n)) ` ` `  `# This code is contributed by ANKITRAI1 `

## C#

 `// C# program to find minimum  ` `// number of operations required  ` `// to make all array elements zero  ` `class` `GFG  ` `{ ` ` `  `// Function to find minimum number  ` `// of operations required to make  ` `// all array elements zero  ` `static` `int` `minimumChanges(``int``[] arr,  ` `                          ``int` `n)  ` `{  ` `    ``int` `i;  ` `     `  `    ``// It will maintain total ` `    ``// changes required  ` `    ``int` `changes = 0;  ` `     `  `    ``for` `(i = 0; i < n; i++)  ` `    ``{  ` `        ``// Check for the first 1  ` `        ``if` `(arr[i] == 1)  ` `        ``{  ` `            ``int` `j;  ` `             `  `            ``// Check for number of  ` `            ``// consecutive 1's  ` `            ``for``(j = i + 1; j < n; j++)  ` `            ``{  ` `                ``if``(arr[j] == 0)  ` `                    ``break``;  ` `            ``}  ` `             `  `            ``// Increment i to the position  ` `            ``// of last consecutive 1  ` `            ``i = j - 1;  ` `             `  `            ``changes++;  ` `        ``}  ` `    ``}  ` `     `  `    ``return` `changes;  ` `}  ` ` `  `// Driver code  ` `static` `void` `Main() ` `{ ` `    ``int``[] arr = ``new` `int``[]{ 1, 1, 0, 0, 0,  ` `                           ``1, 0, 1, 1 };  ` `    ``int` `n = arr.Length ; ` `     `  `    ``System.Console.WriteLine(``"Minimum operations: "` `+ ` `                             ``minimumChanges(arr, n));  ` `} ` `} ` ` `  `// This code is contributed by mits `

## PHP

 `

Output:

```Minimum operations: 3
```

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