Two Sum Less Than K - Practice Coding Problems

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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Understanding the Visualization

Input

Array [34,23,1,24,75,33,54,8] and k=60

Process

Find all pairs where sum < k, keep maximum

Output

Return 58 (maximum sum less than 60)

Key Takeaway

🎯 Key Insight: Sort first to enable efficient two-pointer technique for finding optimal pairs

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

Approach	Time	Space	Notes
✓ Sort + Binary Search	O(n log n)	O(1)	Sort array, then for each element use binary search to find best partner
Brute Force	O(n²)	O(1)	Check all possible pairs to find maximum sum less than k
Sort + Two Pointers	O(n log n)	O(1)	Sort array first, then use two pointers to efficiently find maximum sum

Sort + Binary Search — Algorithm Steps

Sort the array
For each element nums[i], binary search for largest nums[j] where nums[i] + nums[j] < k
Track the maximum sum found

Visualization

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Step-by-Step Walkthrough

Sort

Sort array to enable binary search

Binary Search

For each element, binary search for best partner

Result

Return maximum valid sum found

Code -

solution.c — C

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

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int solution(int* nums, int numsSize, int k) {
    qsort(nums, numsSize, sizeof(int), compare);
    int maxSum = -1;
    
    for (int i = 0; i < numsSize - 1; i++) {
        int target = k - nums[i] - 1;
        int left = i + 1, right = numsSize - 1;
        int bestJ = -1;
        
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] <= target) {
                bestJ = mid;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        if (bestJ != -1) {
            int sum = nums[i] + nums[bestJ];
            if (maxSum == -1 || sum > maxSum) {
                maxSum = sum;
            }
        }
    }
    
    return maxSum;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    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(n log n)

Sorting takes O(n log n), then n binary searches each taking O(log n)

⚡ Linearithmic

Space Complexity

O(1)

Only using a few variables for binary search bounds

✓ Linear Space

23.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Sort + Binary Search — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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