Find the number of zeroes

23 Jun, 2022
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
Approach 1: 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).
Implementation of above approach is below:
C++
#include <bits/stdc++.h>using namespace std;int firstzeroindex(int arr[], int n){    for (int i = 0; i < n; i++) {        if (arr[i] == 0) {            return i;        }    }    return -1;}int main(){    int arr[] = { 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0 };    int n = sizeof(arr) / sizeof(arr[0]);    int x = firstzeroindex(arr, n);    if (x == -1) {        cout << "Count of zero is 0" << endl;    }    else {        cout << "count of zero is " << n - x << endl;    }    return 0;}// this code is contributed by machhaliya muhammad
count of zero is 6
Time complexity: O(n) where n is size of arr.
Approach 2: 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.
Implementation:
C
C++
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Javascript
// A divide and conquer solution to find count of zeroes in an array// where all 1s are present before all 0s#include <stdio.h> /* if 0 is present in arr[] then returns the index of FIRST occurrenceof 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[0]);    printf("Count of zeroes is %d", countZeroes(arr, n));    return 0;}
// A divide and conquer solution to// find count of zeroes in an array// where all 1s are present before all 0s#include <bits/stdc++.h>using namespace std; /* if 0 is present in arr[] thenreturns the index of FIRST occurrenceof 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 Codeint main(){    int arr[] = {1, 1, 1, 0, 0, 0, 0, 0};    int n = sizeof(arr) / sizeof(arr[0]);    cout << "Count of zeroes is "         << countZeroes(arr, n);    return 0;} // This code is contributed by SoumikMondal
// 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));    }}
# 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 -1def 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 Codearr = [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
// A divide and conquer solution to find// count of zeroes in an array where all// 1s are present before all 0susing 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// 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 */function firstZero($arr, $low, $high){    if ($high >= $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?>
<script>// 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     */    function firstZero(arr , low , high) {        if (high >= low) {                     // Check if mid element is first 0            var mid = low + parseInt((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()    function countZeroes(arr , n)    {             // Find index of first zero in given array        var 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         var arr = [ 1, 1, 1, 0, 0, 0, 0, 0 ];        var n = arr.length;        document.write("Count of zeroes is " + countZeroes(arr, n)); // This code is contributed by gauravrajput1</script>
Count of zeroes is 5
Time Complexity: O(Logn) where n is number of elements in arr[].
nitin mittal
SoumikMondal
GauravRajput1
shaheeneallamaiqbal
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