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Contains Duplicate III in C++
Suppose we have an array of integers, we have to check whether there are two distinct indices i and j in the array such that the absolute difference between nums[i] and nums[j] is at most t. And the absolute difference between i and j is at most k. So if input is like [1,2,3,1], then if k = 3 and t = 0, then return true.
To solve this, we will follow these steps −
Make a set s, n := size of nums array
for i in range 0 to n – 1
x is index of set element starting from nums[i] and above
if x is not in range of the set and value of x <= nums[i] + t, then return true
if x is not the first element
x := next element as random
if t th element starting from x is >= nums[i], then return true
insert nums[i] into s, then delete nums[i - k] from s
return false
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
multiset <int> s;
int n = nums.size();
for(int i = 0; i< n; i++){
multiset <int> :: iterator x = s.lower_bound(nums[i]);
if(x != s.end() && *x <= nums[i] + t ) return true;
if(x != s.begin()){
x = std::next(x, -1);
if(*x + t >= nums[i])return true;
}
s.insert(nums[i]);
if(i >= k){
s.erase(nums[i - k]);
}
}
return false;
}
};
main(){
Solution ob;
vector<int> v = {1,2,3,1};
cout << (ob.containsNearbyAlmostDuplicate(v, 3,0));
}Input
[1,2,3,1] 3 0
Output
1
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