Coding Problem Solving

Find pair that sums up to `k`

Senthil Nayagan
Senthil Nayagan           

Problem: Find pair that sums up to “k”


Given an array of integers nums and an integer k, create a boolean function that checks if there are two elements in nums such that we get k when we add them together.

Example 1

  • Input: nums = [4, 1, 5, -3, 6], k = 11
  • Output: true
  • Explanation: 5 + 6 is equivalent to 11

Example 2

  • Input: nums = [4, 1, 5, -3, 6], k = -2
  • Output: true
  • Explanation: 1 + (-3) is equivalent to -2

Solution 1 (Brute force solution)

This solution follows brute force approach.

# Brute-force approach
def find_pair(nums, k):
    for i in range(len(nums)):
        for j in range(i+1, len(nums)):
            if nums[i] + nums[j] == k:
                return True
    return False

if __name__ == "__main__":
    nums = [4, 1, 5, -3, 6]

    print(find_pair(nums, 11)) # True
    print(find_pair(nums, -4)) # False
    print(find_pair(nums, -2)) # True


  • Time complexity: O(n2)
  • Space complexity: O(1)

Let’s find a better solution than this one.

Solution 2

This approach begins by sorting the numbers and then reduces the amount of traversal needed. Since the numbers are sorted in ascending order, then:

  • nums[i] >= nums[i-1]
  • nums[i] <= nums[i+1]

This approach uses the left index and the right index. If we increasing the left index, the sum value (k) will either increase or remain the same. In a similar manner, decreasing the right index will either bring about a reduction in the sum value (k) or cause it to stay unchanged.

def find_pair(nums, k):
    left_idx = 0
    right_idx = len(nums) - 1

    while left_idx < right_idx:
        if nums[left_idx] + nums[right_idx] == k:
            return True
        elif nums[left_idx] + nums[right_idx] < k:
            left_idx += 1
            right_idx -= 1
    return False


  • Time complexity: O(n log n)
  • Space complexity: Depends on the sorting algorithm we use. For example, if it’s O(log n), then the space complexity of this algorithm is O(long n).

Let’s find a better solution than this one.

Solution 3 (Using hash table. It’s the most optimal solution)

This solution uses hash table. The hash table is a powerful tool when solving coding problems because it has an O(1) lookup on average, so we can get the value of a certain key in O(1). Also, it has an O(1) insertion on average, so we can insert an element in O(1).

def find_pair(nums, k):
    visited = {}  # Dictionary as hash table

    for element in nums:
        if visited.get(k - element):  # O(1) for lookup
            return True
            visited[element] = True  # O(1) for insertion
    return False


  • Time complexity: O(n)
  • Space complexity: O(n) - We are using additional space for a hash table that can contain n elements in the worst case.

The lookup and insertion are constant on average in this case. Hence, the O(n).


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