Two Sum Less Than K

Two Sum Less Than K - Problem

Given an array nums of integers and integer k, return the maximum sum such that there exists i < j with nums[i] + nums[j] = sum and sum < k.

If no i, j exist satisfying this equation, return -1.

Input & Output

Example 1 — Basic Case

$ Input: nums = [34,23,1,24,75,33,54,8], k = 60

› Output: 58

💡 Note: We can choose the pair (1, 54) or (24, 34) both giving sum 58, which is the maximum sum less than 60.

Example 2 — No Valid Pairs

$ Input: nums = [10,20,30], k = 15

› Output: -1

💡 Note: The smallest possible sum is 10 + 20 = 30, which is not less than k = 15, so return -1.

Example 3 — Small Numbers

$ Input: nums = [1,2,3,4], k = 5

› Output: 4

💡 Note: Choose pair (1, 3) with sum 4, which is the maximum sum less than k = 5.

Constraints

2 ≤ nums.length ≤ 10³
1 ≤ nums[i] ≤ 1000
1 ≤ k ≤ 2000

Visualization

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Asked in

a Amazon 15 f Facebook 8 G Google 12

The key insight is to find pairs efficiently by sorting first, then using two pointers or binary search. The optimal approach is sort + two pointers with Time: O(n log n), Space: O(1).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

Iterate through every possible pair of elements and track the maximum sum that is less than k. For each element at index i, check all elements at indices j > i.

Sort + Binary Search

⏱️ Time: O(n log n) Space: O(1)

Sort the array first. For each element at index i, use binary search to find the largest element at index j > i such that nums[i] + nums[j] < k.

Sort + Two Pointers

⏱️ Time: O(n log n) Space: O(1)

Sort the array, then use two pointers (left and right) to find pairs. If sum < k, try to increase it by moving left pointer. If sum ≥ k, decrease it by moving right pointer.

Brute Force — Algorithm Steps

Initialize maxSum to -1
Use nested loops to check all pairs (i, j) where i < j
If nums[i] + nums[j] < k, update maxSum if this sum is larger

Visualization

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Step-by-Step Walkthrough

Setup

Initialize maxSum = -1, check all pairs

Compare

For each pair, check if sum < k and update maximum

Result

Return the maximum valid sum found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize, int k) {
    int maxSum = -1;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            int currentSum = nums[i] + nums[j];
            if (currentSum < k) {
                if (maxSum == -1 || currentSum > maxSum) {
                    maxSum = currentSum;
                }
            }
        }
    }
    
    return maxSum;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n' && token[0] != ']') {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, numsSize, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all n*(n-1)/2 possible pairs

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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