# Contains Duplicate III in C++

C++Server Side ProgrammingProgramming

#### C in Depth: The Complete C Programming Guide for Beginners

45 Lectures 4.5 hours

#### Practical C++: Learn C++ Basics Step by Step

Most Popular

50 Lectures 4.5 hours

#### Master C and Embedded C Programming- Learn as you go

Best Seller

66 Lectures 5.5 hours

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 −

Live Demo

#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
Updated on 02-May-2020 07:11:25