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You are given an integer array `nums`

and an integer `k`

.

In one operation, you can pick two numbers from the array whose sum equals `k`

and remove them from the array.

Return *the maximum number of operations you can perform on the array*.

**Example 1:**

Input:nums = [1,2,3,4], k = 5Output:2Explanation:Starting with nums = [1,2,3,4]: - Remove numbers 1 and 4, then nums = [2,3] - Remove numbers 2 and 3, then nums = [] There are no more pairs that sum up to 5, hence a total of 2 operations.

**Example 2:**

Input:nums = [3,1,3,4,3], k = 6Output:1Explanation:Starting with nums = [3,1,3,4,3]: - Remove the first two 3's, then nums = [1,4,3] There are no more pairs that sum up to 6, hence a total of 1 operation.

**Constraints:**

`1 <= nums.length <= 10`

^{5}`1 <= nums[i] <= 10`

^{9}`1 <= k <= 10`

^{9}

` ````
```class Solution {
public:
int maxOperations(vector& nums, int k) {
sort(nums.begin(),nums.end());
if(nums.size()<2)
return 0;
int left=0;
int right=nums.size()-1;
int count=0;
while(leftk)
right--;
else
{
count++;
left++;
right--;
}
}
return count;
}
};

In this program we have to find the maximum number of pairs of elements in the vector ‘nums’ whose sum is equal to given target ‘k’ .

We will do this by using “two-pointer approach”.

Let’s see the step-by-step approach of the code:

- First of all the array should be in a sorted form so, we sort the ‘nums’ in ascending order using ‘sort(nums.begin( ) , nums.end( ) ) ‘ this is a very important step for the two- pointer approach to work efficiently.

- Next we check the size of the vector , if the size of the vector is less than 2 , it means no pairs or groups can be formed and the function returns 0.

- Now we initializes two pointers , ‘left’ and ‘right’ , where left points to the beginning of the sorted array and right points towards the end .

- After that, we initializes a variable ‘count’ to 0 , for tracking the number of pairs found.

- ‘While’ loop has been used here in which the loop will continue till the point when ‘left’ is less than ‘right’

In this loop :

- if the sum of of elements of positions of ‘left’ and ‘right’ is less than the target ‘k’ .

- if the sum is less, it means we need a larger sum , so we simply increment ‘left’ .

- if the the sum is greater, that means we need a smaller sum so we decrements ‘right’ .

- if the the sum is equal to ‘k’ , it means the pair is found, and we increment the count ( count ++) and move both pointers (‘left++’ and ’right- -‘ ) so that we don’t count these elements again.

- The loop will continue until ‘left’ is no longer less than ‘right’.

- Finally, the function returns the total count pairs found .

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