Description#
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 <= 105nums.length == n + 11 <= nums[i] <= n- All the integers in
nums appear only once except for precisely one integer which appears two or more times.
Follow up:
- How can we prove that at least one duplicate number must exist in
nums? - Can you solve the problem in linear runtime complexity?
Solutions#
Solution 1: Binary Search#
We can observe that if the number of elements in $[1,..x]$ is greater than $x$, then the duplicate number must be in $[1,..x]$, otherwise the duplicate number must be in $[x+1,..n]$.
Therefore, we can use binary search to find $x$, and check whether the number of elements in $[1,..x]$ is greater than $x$ at each iteration. This way, we can determine which interval the duplicate number is in, and narrow down the search range until we find the duplicate number.
The time complexity is $O(n \times \log n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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| class Solution:
def findDuplicate(self, nums: List[int]) -> int:
def f(x: int) -> bool:
return sum(v <= x for v in nums) > x
return bisect_left(range(len(nums)), True, key=f)
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| class Solution {
public int findDuplicate(int[] nums) {
int l = 0, r = nums.length - 1;
while (l < r) {
int mid = (l + r) >> 1;
int cnt = 0;
for (int v : nums) {
if (v <= mid) {
++cnt;
}
}
if (cnt > mid) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}
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| class Solution {
public:
int findDuplicate(vector<int>& nums) {
int l = 0, r = nums.size() - 1;
while (l < r) {
int mid = (l + r) >> 1;
int cnt = 0;
for (int& v : nums) {
cnt += v <= mid;
}
if (cnt > mid) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
};
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| func findDuplicate(nums []int) int {
return sort.Search(len(nums), func(x int) bool {
cnt := 0
for _, v := range nums {
if v <= x {
cnt++
}
}
return cnt > x
})
}
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| function findDuplicate(nums: number[]): number {
let l = 0;
let r = nums.length - 1;
while (l < r) {
const mid = (l + r) >> 1;
let cnt = 0;
for (const v of nums) {
if (v <= mid) {
++cnt;
}
}
if (cnt > mid) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
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| impl Solution {
#[allow(dead_code)]
pub fn find_duplicate(nums: Vec<i32>) -> i32 {
let mut left = 0;
let mut right = nums.len() - 1;
while left < right {
let mid = (left + right) >> 1;
let cnt = nums
.iter()
.filter(|x| **x <= (mid as i32))
.count();
if cnt > mid {
right = mid;
} else {
left = mid + 1;
}
}
left as i32
}
}
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| /**
* @param {number[]} nums
* @return {number}
*/
var findDuplicate = function (nums) {
let l = 0;
let r = nums.length - 1;
while (l < r) {
const mid = (l + r) >> 1;
let cnt = 0;
for (const v of nums) {
if (v <= mid) {
++cnt;
}
}
if (cnt > mid) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
|
