Java Program to Count 1's in a sorted binary array Last Updated : 27 Dec, 2021 Comments Improve Suggest changes Like Article Like Report 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. 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 code */ 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 */ OutputCount of 1's in given array is 4 Time complexity of the above solution is O(Logn) Space complexity o(log n) (function call stack) The same approach with iterative solution would be Java /*package whatever //do not write package name here */ import java.io.*; class GFG { static int countOnes(int arr[], int n) { int ans; int low = 0, high = n - 1; while (low <= high) { // get the middle index int mid = (low + high) / 2; // else recur for left side if (arr[mid] < 1) high = mid - 1; // If element is not last 1, recur for right side else if (arr[mid] > 1) low = mid + 1; else // check if the element at middle index is last 1 { if (mid == n - 1 || arr[mid + 1] != 1) return mid + 1; else low = mid + 1; } } return 0; } // Driver code public static void main (String[] args) { int arr[] = { 1, 1, 1, 1, 0, 0, 0 }; int n = arr.length; System.out.println("Count of 1's in given array is "+ countOnes(arr, n)); } } // This code is contributed by patel2127. OutputCount of 1's in given array is 4 Time complexity of the above solution is O(Logn) Space complexity is O(1) Please refer complete article on Count 1's in a sorted binary array for more details! Comment More infoAdvertise with us Next Article Java Program to Count 1's in a sorted binary array kartik Follow Improve Article Tags : Java Searching Java Programs DSA Binary Search binary-string +2 More Practice Tags : Binary SearchJavaSearching Similar Reads Java Program for Counting sets of 1s and 0s in a binary matrix Given a n × m binary matrix, count the number of sets where a set can be formed one or more same values in a row or column. Examples: Input: 1 0 1 0 1 0 Output: 8 Explanation: There are six one-element sets (three 1s and three 0s). There are two two- element sets, the first one consists of the first 3 min read Java Program to Convert a Decimal Number to Binary Number using Arrays as Stacks Given an Integer number convert into Binary Number using arrays as a stack. 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