Counting Elements LeetCode | Counting Elements LeetCode Java | Programming Tutorials

Counting Elements LeetCode Solution using Java Code.

Given an integer array, count element x such that x + 1 is also in array.If there're duplicates in array, count them separately.

Example 1:

Input: {1, 2, 3}
Output: 2

Explanation:

First element is 1 + 1 = 2 ( 2 is present in an array)
Second element is 2 + 1 = 3 ( 3 is present in an array)
Third element is 3 + 1 = 4 ( 4 is not present in an array)


Example 2:

Input: {1, 1, 3, 3, 5, 5, 7, 7}
Output: 0

Example 3:

Input : {1, 3, 2, 3, 5, 0}
Output: 3

Explanation:
1 + 1 = 2 (Exist)
3 + 1 = 4 (Not exist)
2 + 1 = 3 (Exist)
3 + 1 = 4 (Not exist)
5 + 1 = 6 (Not exist)
0 + 1 = 1 (Exist)

Example 4:

Input : {1, 1, 2, 2}
Output: 2

This problem is the day 7 challenge of LeetCode 30 day challenge.

[Neilblaze]

Neat and to the point explanation!
Thanks Sir :)

[naveenverma]

Good explanation brother tnx keep doing this

[blindprogrammer]

Thanks for your video! Following is the solution in C++
int countElements(vector<int>& arr)
{
unordered_set<int> st;
vector<int>::iterator ptr;
unordered_set<int>::iterator s;
int count = 0;

for(ptr = arr.begin(); ptr != arr.end(); ptr++)
st.insert(*ptr);

for(ptr = arr.begin(); ptr != arr.end(); ptr++)
if (st.find(*ptr + 1) != st.end())
count++;

return count;
}

[michaellieberman]

I'm trying to think of a faster way which uses only one run through of the array (O(n)).

Create HashMap of <Integer we've seen> <Integer's which complement integers we've seen>
for (int i = 0; i < arr.length; i++){
if (hashMap.contains(arr[i])){
total++;
}
hashMap.add(arr[i], arr[i] + 1);
hashMap.add(arr[i], arr[i] - 1);

}


return total;


of course if arr[i] is already part of the keys, prob dont need to add to save space.

maybe just do <integer we've seen> <array of integers which complement integers we've seen> to avoid adding twice

[chomuekez]

how do you validate against empty input, to avoid format exception

[omarz]

Can you help me figure out why my code is not working in javascript?

function countElements(arr) {
let set = new Set()
let count = 0

for(let i = 0; i< arr.length; i++){
set.add(arr)
}
for(let i = 0; i < arr.length; i++){
if (set.has(arr[i] +1)){
count++
}
}
return count
}

[adityabej]

Can we not use stacks for implementation ?

[amiteshrajvaidya]

no sir this code does not works for all the cases.. it cant pass the testcase of [1, 1, 2] . according to this code the answer would be 2 but the real answer is 1 for this.

[Akashi]

C# Code
//Runtime:92 ms
//Memory Usage: 23.9 MB
public int CountElements(int[] arr)
{
int counter = 0;
for (int i = 0; i < arr.Length; i++)
{ int j = 0;
while(j< arr.Length)
{
if(i!=j&& arr[i]+1== arr[j])
{
counter++;
break;
}
j++;
}
}
return counter;
}

[rahullotlikar]

It does not work for the case arr =[1, 3, 2, 3, 5, 0] as ther are two 3's