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.
Example 1:
- Input: [1,3,4,2,2]
- Output: 2
- Input: [3,1,3,4,2]
- Output: 3
- 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(n^2).
- There is only one duplicate number in the array, but it could be repeated more than once.
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Solution:
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/** | |
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. | |
Example 1: | |
Input: [1,3,4,2,2] | |
Output: 2 | |
Example 2: | |
Input: [3,1,3,4,2] | |
Output: 3 | |
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(n^2). | |
There is only one duplicate number in the array, but it could be repeated more than once. | |
*/ | |
public class ArraySingleDuplicate{ | |
/** | |
-use the concept of binary search | |
*/ | |
public static int findDuplicate(int[] nums) { | |
int low = 1, high = nums.length-1; | |
int mid = 0; | |
while(low<high){ | |
int count = 0; | |
mid = (low+high)/2; | |
for(int i:nums){ | |
if(i<=mid){ | |
count++; | |
} | |
} | |
if(count<=mid){//because the expected elements on the left to mid are less than or equal to mid | |
low = mid+1; | |
}else{ | |
high = mid; | |
} | |
} | |
return low; | |
} | |
public static void main(String args[]){ | |
int nums [] = new int[]{1,3,4,2,2}; | |
System.out.println("dup is:"+findDuplicate(nums)); | |
} | |
} |
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