Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Note:
- You must not modify the array (assume the array is read only).
- You must use only constant, O(1) extra space.
- Your runtime complexity should be less than
O(n2)
. - There is only one duplicate number in the array, but it could be repeated more than once.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
Analysis:
if duplicated number is less than X, then there must be more than X numbers in the array less than X.
So, we use binary search from 1 to n, count the number of number <= mid, and determine whether is the duplicated number.
Solution:
public class Solution { public int findDuplicate(int[] nums) { if (nums.length == 0) return 0; int start = 1, end = nums.length - 1; // start is 1, end is n. while (start <= end) { int mid = start + (end - start) / 2; // count the numbers no larger than mid. int count = 0; for (int num : nums) if (num <= mid) { count++; } if (count > mid) { end = mid - 1; } else { start = mid + 1; } } return start; } }