Given a sorted array of length n, find the number in array that appears more than or equal to n/2 times. It is given that such element always exists.
Input : 2 3 3 4 Output : 3 Input : 3 4 5 5 5 Output : 5 Input : 1 1 1 2 3 Output : 1
To find that number, we traverse the array and check the frequency of every element in array if it is greater than or equals to n/2 but it requires extra space and time complexity will be O(n).
But we can see that the if there is number that comes more than or equal to n/2 times in a sorted array, then that number must be present at the position n/2 i.e. a[n/2].
Time Complexity : O(1)
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