- Contains Duplicate III
Hard
You are given an integer array nums and two integers indexDiff and valueDiff.
Find a pair of indices (i, j) such that:
i != j,
abs(i - j) <= indexDiff.
abs(nums[i] - nums[j]) <= valueDiff, and
Return true if such pair exists or false otherwise.
Example 1:
Input: nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
Output: true
Explanation: We can choose (i, j) = (0, 3).
We satisfy the three conditions:
i != j --> 0 != 3
abs(i - j) <= indexDiff --> abs(0 - 3) <= 3
abs(nums[i] - nums[j]) <= valueDiff --> abs(1 - 1) <= 0
Example 2:
Input: nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
Output: false
Explanation: After trying all the possible pairs (i, j), we cannot satisfy the three conditions, so we return false.
Constraints:
2 <= nums.length <= 105
-109 <= nums[i] <= 109
1 <= indexDiff <= nums.length
0 <= valueDiff <= 109
Solution 1: This question requires using the lower_bound function of set, which means the smallest element in the set >= the input value.
class Solution {
public:
bool containsNearbyAlmostDuplicate(vector<int>& nums, int indexDiff, int valueDiff) {
int n = nums.size();
int left = 0, right = 0;
// unordered_map<int, int> mp; //<value, index> 注意用map不对，因为要检测map里面所有的entry，时间复杂度高。
set<int> s;
while (right < n) {
//mp[nums[right]] = right;
auto iter = s.lower_bound(nums[right]);
if (iter != s.end()) {
if (*iter - nums[right] <= valueDiff) return true;
}
iter = s.lower_bound(nums[right]);
if (iter != s.begin()) {
iter--;
if (nums[right] - *iter <= valueDiff) return true;
}
s.insert(nums[right]);
right++;
while (right - left > indexDiff) {
s.erase(nums[left]);
left++;
}
}
return false;
}
};