Array Single Duplicate LeetCode (Find the Duplicate Number)

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.

This problem is an easy problem and is popular in LeetCode and other forums - Link 1 A collection of hundreds of interview questions and solutions are available in our blog at Interview Question Solutions

Solution:

 

/**
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));
}
}

No comments:

Post a Comment