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# Find the number of zeroes

• Difficulty Level : Easy
• Last Updated : 02 Jun, 2020

Given an array of 1s and 0s which has all 1s first followed by all 0s. Find the number of 0s. Count the number of zeroes in the given array.

Examples :

```Input: arr[] = {1, 1, 1, 1, 0, 0}
Output: 2

Input: arr[] = {1, 0, 0, 0, 0}
Output: 4

Input: arr[] = {0, 0, 0}
Output: 3

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

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution is to traverse the input array. As soon as we find a 0, we return n – index of first 0. Here n is number of elements in input array. Time complexity of this solution would be O(n).

Since the input array is sorted, we can use Binary Search to find the first occurrence of 0. Once we have index of first element, we can return count as n – index of first zero.

## C

 `// A divide and conquer solution to find count of zeroes in an array``// where all 1s are present before all 0s``#include `` ` `/* if 0 is present in arr[] then returns the index of FIRST occurrence``   ``of 0 in arr[low..high], otherwise returns -1 */``int` `firstZero(``int` `arr[], ``int` `low, ``int` `high)``{``    ``if` `(high >= low)``    ``{``        ``// Check if mid element is first 0``        ``int` `mid = low + (high - low)/2;``        ``if` `(( mid == 0 || arr[mid-1] == 1) && arr[mid] == 0)``            ``return` `mid;`` ` `        ``if` `(arr[mid] == 1)  ``// If mid element is not 0``            ``return` `firstZero(arr, (mid + 1), high);``        ``else`  `// If mid element is 0, but not first 0``            ``return` `firstZero(arr, low, (mid -1));``    ``}``    ``return` `-1;``}`` ` `// A wrapper over recursive function firstZero()``int` `countZeroes(``int` `arr[], ``int` `n)``{``    ``// Find index of first zero in given array``    ``int` `first = firstZero(arr, 0, n-1);`` ` `    ``// If 0 is not present at all, return 0``    ``if` `(first == -1)``        ``return` `0;`` ` `    ``return` `(n - first);``}`` ` `/* Driver program to check above functions */``int` `main()``{``    ``int` `arr[] =   {1, 1, 1, 0, 0, 0, 0, 0};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr);``    ``printf``(``"Count of zeroes is %d"``, countZeroes(arr, n));``    ``return` `0;``}`

## C++

 `// A divide and conquer solution to ``// find count of zeroes in an array``// where all 1s are present before all 0s``#include ``using` `namespace` `std;`` ` `/* if 0 is present in arr[] then``returns the index of FIRST occurrence``of 0 in arr[low..high], otherwise returns -1 */``int` `firstZero(``int` `arr[], ``int` `low, ``int` `high)``{``    ``if` `(high >= low)``    ``{``        ``// Check if mid element is first 0``        ``int` `mid = low + (high - low) / 2;``        ``if` `((mid == 0 || arr[mid - 1] == 1) && ``                         ``arr[mid] == 0)``            ``return` `mid;`` ` `        ``// If mid element is not 0``        ``if` `(arr[mid] == 1) ``            ``return` `firstZero(arr, (mid + 1), high);``         ` `        ``// If mid element is 0, but not first 0    ``        ``else` `            ``return` `firstZero(arr, low, (mid -1));``    ``}``    ``return` `-1;``}`` ` `// A wrapper over recursive function firstZero()``int` `countZeroes(``int` `arr[], ``int` `n)``{``    ``// Find index of first zero in given array``    ``int` `first = firstZero(arr, 0, n - 1);`` ` `    ``// If 0 is not present at all, return 0``    ``if` `(first == -1)``        ``return` `0;`` ` `    ``return` `(n - first);``}`` ` `// Driver Code``int` `main()``{``    ``int` `arr[] = {1, 1, 1, 0, 0, 0, 0, 0};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << ``"Count of zeroes is "``         ``<< countZeroes(arr, n);``    ``return` `0;``}`` ` `// This code is contributed by SoumikMondal`

## Java

 `// A divide and conquer solution to find count of zeroes in an array``// where all 1s are present before all 0s`` ` `class` `CountZeros ``{``    ``/* if 0 is present in arr[] then returns the index of FIRST occurrence``       ``of 0 in arr[low..high], otherwise returns -1 */``    ``int` `firstZero(``int` `arr[], ``int` `low, ``int` `high) ``    ``{``        ``if` `(high >= low) ``        ``{``            ``// Check if mid element is first 0``            ``int` `mid = low + (high - low) / ``2``;``            ``if` `((mid == ``0` `|| arr[mid - ``1``] == ``1``) && arr[mid] == ``0``)``                ``return` `mid;`` ` `            ``if` `(arr[mid] == ``1``) ``// If mid element is not 0``                ``return` `firstZero(arr, (mid + ``1``), high);``            ``else` `// If mid element is 0, but not first 0``                ``return` `firstZero(arr, low, (mid - ``1``));``        ``}``        ``return` `-``1``;``    ``}`` ` `    ``// A wrapper over recursive function firstZero()``    ``int` `countZeroes(``int` `arr[], ``int` `n) ``    ``{``        ``// Find index of first zero in given array``        ``int` `first = firstZero(arr, ``0``, n - ``1``);`` ` `        ``// If 0 is not present at all, return 0``        ``if` `(first == -``1``)``            ``return` `0``;`` ` `        ``return` `(n - first);``    ``}`` ` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args) ``    ``{``        ``CountZeros count = ``new` `CountZeros();``        ``int` `arr[] = {``1``, ``1``, ``1``, ``0``, ``0``, ``0``, ``0``, ``0``};``        ``int` `n = arr.length;``        ``System.out.println(``"Count of zeroes is "` `+ count.countZeroes(arr, n));``    ``}``}`

## Python3

 `# A divide and conquer solution to``# find count of zeroes in an array``# where all 1s are present before all 0s`` ` `# if 0 is present in arr[] then returns``# the index of FIRST occurrence of 0 in``# arr[low..high], otherwise returns -1 ``def` `firstZero(arr, low, high):`` ` `    ``if` `(high >``=` `low):``     ` `        ``# Check if mid element is first 0``        ``mid ``=` `low ``+` `int``((high ``-` `low) ``/` `2``)``        ``if` `(( mid ``=``=` `0` `or` `arr[mid``-``1``] ``=``=` `1``)``                      ``and` `arr[mid] ``=``=` `0``):``            ``return` `mid``         ` `        ``# If mid element is not 0``        ``if` `(arr[mid] ``=``=` `1``): ``            ``return` `firstZero(arr, (mid ``+` `1``), high)``             ` `        ``# If mid element is 0, but not first 0``        ``else``: ``            ``return` `firstZero(arr, low, (mid ``-` `1``))``     ` `    ``return` `-``1`` ` `# A wrapper over recursive``# function firstZero()``def` `countZeroes(arr, n):`` ` `    ``# Find index of first zero in given array``    ``first ``=` `firstZero(arr, ``0``, n ``-` `1``)`` ` `    ``# If 0 is not present at all, return 0``    ``if` `(first ``=``=` `-``1``):``        ``return` `0`` ` `    ``return` `(n ``-` `first)`` ` `# Driver Code``arr ``=` `[``1``, ``1``, ``1``, ``0``, ``0``, ``0``, ``0``, ``0``]``n ``=` `len``(arr)``print``(``"Count of zeroes is"``,``        ``countZeroes(arr, n))`` ` `# This code is contributed by Smitha Dinesh Semwal`

## C#

 `// A divide and conquer solution to find ``// count of zeroes in an array where all``// 1s are present before all 0s``using` `System;`` ` `class` `CountZeros ``{``    ``/* if 0 is present in arr[] then returns``       ``the index of FIRST occurrence of 0 in``       ``arr[low..high], otherwise returns -1 */``    ``int` `firstZero(``int` `[]arr, ``int` `low, ``int` `high) ``    ``{``        ``if` `(high >= low) ``        ``{``            ``// Check if mid element is first 0``            ``int` `mid = low + (high - low) / 2;``            ``if` `((mid == 0 || arr[mid - 1] == 1) &&``                                 ``arr[mid] == 0)``                ``return` `mid;`` ` `            ``if` `(arr[mid] == 1) ``// If mid element is not 0``                ``return` `firstZero(arr, (mid + 1), high);``                 ` `            ``else` `// If mid element is 0, but not first 0``                ``return` `firstZero(arr, low, (mid - 1));``        ``}``        ``return` `-1;``    ``}`` ` `    ``// A wrapper over recursive function firstZero()``    ``int` `countZeroes(``int` `[]arr, ``int` `n) ``    ``{``        ``// Find index of first zero in given array``        ``int` `first = firstZero(arr, 0, n - 1);`` ` `        ``// If 0 is not present at all, return 0``        ``if` `(first == -1)``            ``return` `0;`` ` `        ``return` `(n - first);``    ``}`` ` `    ``// Driver program to test above functions``    ``public` `static` `void` `Main() ``    ``{``        ``CountZeros count = ``new` `CountZeros();``        ``int` `[]arr = {1, 1, 1, 0, 0, 0, 0, 0};``        ``int` `n = arr.Length;``        ``Console.Write(``"Count of zeroes is "` `+ ``                       ``count.countZeroes(arr, n));``    ``}``}`` ` `// This code is contributed by nitin mittal.`

## PHP

 `= ``\$low``)``    ``{``         ` `        ``// Check if mid element is first 0``        ``\$mid` `= ``\$low` `+ ``floor``((``\$high` `- ``\$low``)/2);``         ` `        ``if` `(( ``\$mid` `== 0 || ``\$arr``[``\$mid``-1] == 1) && ``                                 ``\$arr``[``\$mid``] == 0)``            ``return` `\$mid``;`` ` `        ``// If mid element is not 0``        ``if` `(``\$arr``[``\$mid``] == 1) ``            ``return` `firstZero(``\$arr``, (``\$mid` `+ 1), ``\$high``);``         ` `        ``// If mid element is 0,``        ``// but not first 0    ``        ``else``            ``return` `firstZero(``\$arr``, ``\$low``, ``                            ``(``\$mid` `- 1));``    ``}``    ``return` `-1;``}`` ` `// A wrapper over recursive``// function firstZero()``function` `countZeroes(``\$arr``, ``\$n``)``{``     ` `    ``// Find index of first ``    ``// zero in given array``    ``\$first` `= firstZero(``\$arr``, 0, ``\$n` `- 1);`` ` `    ``// If 0 is not present``    ``// at all, return 0``    ``if` `(``\$first` `== -1)``        ``return` `0;`` ` `    ``return` `(``\$n` `- ``\$first``);``}``     ` `    ``// Driver Code``    ``\$arr` `= ``array``(1, 1, 1, 0, 0, 0, 0, 0);``    ``\$n` `= sizeof(``\$arr``);``    ``echo``(``"Count of zeroes is "``); ``    ``echo``(countZeroes(``\$arr``, ``\$n``));``     ` `// This code is contributed by nitin mittal``?>`

Output:

`Count of zeroes is 5 `

Time Complexity: O(Logn) where n is number of elements in arr[].