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# Find the Duplicate Number LeetCode Solution

## Problem – Find the Duplicate Number

Given an array of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.

There is only one repeated number in `nums`, return this repeated number.

You must solve the problem without modifying the array `nums` and uses only constant extra space.

Example 1:

``````Input: nums = [1,3,4,2,2]
Output: 2
``````

Example 2:

``````Input: nums = [3,1,3,4,2]
Output: 3``````

Constraints:

• `1 <= n <= 105`
• `nums.length == n + 1`
• `1 <= nums[i] <= n`
• All the integers in `nums` appear only once except for precisely one integer which appears two or more times.

• How can we prove that at least one duplicate number must exist in `nums`?
• Can you solve the problem in linear runtime complexity?

### Find the Duplicate Number LeetCode Solution in Java

``````    public static int findDuplicate_2loops(int[] nums) {
int len = nums.length;
for (int i = 0; i < len; i++) {
for (int j = i + 1; j < len; j++) {
if (nums[i] == nums[j]) {
return nums[i];
}
}
}

return len;
}
``````

### Find the Duplicate Number LeetCode Solution in C++

``````class Solution {
public:
int findDuplicate(vector<int>& nums) {
int low = 1, high = nums.size() - 1, cnt;

while(low <=  high)
{
int mid = low + (high - low) / 2;
cnt = 0;
// cnt number less than equal to mid
for(int n : nums)
{
if(n <= mid)
++cnt;
}
// binary search on left
if(cnt <= mid)
low = mid + 1;
else
// binary search on right
high = mid - 1;

}
return low;
}
};
``````

### Find the Duplicate Number LeetCode Solution in Python

``````class Solution:
def findDuplicate(self, nums):
slow, fast = nums[0], nums[0]
while True:
slow, fast = nums[slow], nums[nums[fast]]
if slow == fast: break

slow = nums[0];
while slow != fast:
slow, fast = nums[slow], nums[fast]
return slow
``````
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