Given a binary array sorted in non-increasing order, count the number of 1’s in it.
Examples:
Input: arr[] = {1, 1, 0, 0, 0, 0, 0}
Output: 2
Input: arr[] = {1, 1, 1, 1, 1, 1, 1}
Output: 7
Input: arr[] = {0, 0, 0, 0, 0, 0, 0}
Output: 0
A simple solution is to linearly traverse the array. The time complexity of the simple solution is O(n). We can use Binary Search to find count in O(Logn) time. The idea is to look for last occurrence of 1 using Binary Search. Once we find the index last occurrence, we return index + 1 as count.
The following is the implementation of above idea.
C++
// C++ program to count one's in a boolean array #include <bits/stdc++.h> using namespace std; /* Returns counts of 1's in arr[low..high]. The array is assumed to be sorted in non-increasing order */int countOnes(bool arr[], int low, int high) { if (high >= low) { // get the middle index int mid = low + (high - low)/2; // check if the element at middle index is last 1 if ( (mid == high || arr[mid+1] == 0) && (arr[mid] == 1)) return mid+1; // If element is not last 1, recur for right side if (arr[mid] == 1) return countOnes(arr, (mid + 1), high); // else recur for left side return countOnes(arr, low, (mid -1)); } return 0; } /* Driver program to test above functions */int main() { bool arr[] = {1, 1, 1, 1, 0, 0, 0}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Count of 1's in given array is " << countOnes(arr, 0, n-1); return 0; } |
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Python
# Python program to count one's in a boolean array # Returns counts of 1's in arr[low..high]. The array is # assumed to be sorted in non-increasing order def countOnes(arr,low,high): if high>=low: # get the middle index mid = low + (high-low)/2 # check if the element at middle index is last 1 if ((mid == high or arr[mid+1]==0) and (arr[mid]==1)): return mid+1 # If element is not last 1, recur for right side if arr[mid]==1: return countOnes(arr, (mid+1), high) # else recur for left side return countOnes(arr, low, mid-1) return 0 # Driver function arr=[1, 1, 1, 1, 0, 0, 0] print "Count of 1's in given array is",countOnes(arr, 0 , len(arr)-1) # This code is contributed by __Devesh Agrawal__ |
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Java
// Java program to count 1's in a sorted array class CountOnes { /* Returns counts of 1's in arr[low..high]. The array is assumed to be sorted in non-increasing order */ int countOnes(int arr[], int low, int high) { if (high >= low) { // get the middle index int mid = low + (high - low)/2; // check if the element at middle index is last 1 if ( (mid == high || arr[mid+1] == 0) && (arr[mid] == 1)) return mid+1; // If element is not last 1, recur for right side if (arr[mid] == 1) return countOnes(arr, (mid + 1), high); // else recur for left side return countOnes(arr, low, (mid -1)); } return 0; } /* Driver program to test above functions */ public static void main(String args[]) { CountOnes ob = new CountOnes(); int arr[] = {1, 1, 1, 1, 0, 0, 0}; int n = arr.length; System.out.println("Count of 1's in given array is " + ob.countOnes(arr, 0, n-1) ); } } /* This code is contributed by Rajat Mishra */ |
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C#
// C# program to count 1's in a sorted array using System; class GFG { /* Returns counts of 1's in arr[low..high]. The array is assumed to be sorted in non-increasing order */ static int countOnes(int []arr, int low, int high) { if (high >= low) { // get the middle index int mid = low + (high - low) / 2; // check if the element at middle // index is last 1 if ( (mid == high || arr[mid+1] == 0) && (arr[mid] == 1)) return mid+1; // If element is not last 1, recur // for right side if (arr[mid] == 1) return countOnes(arr, (mid + 1), high); // else recur for left side return countOnes(arr, low, (mid - 1)); } return 0; } /* Driver program to test above functions */ public static void Main() { int []arr = {1, 1, 1, 1, 0, 0, 0}; int n = arr.Length; Console.WriteLine("Count of 1's in given " + "array is " + countOnes(arr, 0, n-1) ); } } // This code is contributed by Sam007. |
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PHP
<?php // PHP program to count one's in a // boolean array /* Returns counts of 1's in arr[low..high]. The array is assumed to be sorted in non-increasing order */function countOnes( $arr, $low, $high) { if ($high >= $low) { // get the middle index $mid = $low + ($high - $low)/2; // check if the element at middle // index is last 1 if ( ($mid == $high or $arr[$mid+1] == 0) and ($arr[$mid] == 1)) return $mid+1; // If element is not last 1, recur for // right side if ($arr[$mid] == 1) return countOnes($arr, ($mid + 1), $high); // else recur for left side return countOnes($arr, $low, ($mid -1)); } return 0; } /* Driver program to test above functions */$arr = array(1, 1, 1, 1, 0, 0, 0); $n = count($arr); echo "Count of 1's in given array is " , countOnes($arr, 0, $n-1); // This code is contributed by anuj_67. ?> |
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Output:
Count of 1's in given array is 4
Time complexity of the above solution is O(Logn)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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