Find the Kth Smallest Sum of a Matrix With Sorted Rows

Find the Kth Smallest Sum of a Matrix With Sorted Rows - Problem

Imagine you're a data analyst tasked with finding specific statistical insights from a structured dataset. You have an m × n matrix where each row is sorted in non-decreasing order, representing different categories of data points.

Your challenge: select exactly one element from each row to form a combination, then find the k-th smallest sum among all possible combinations.

For example, with matrix [[1,3,11],[2,4,6]] and k=5:

Possible combinations: [1,2], [1,4], [1,6], [3,2], [3,4], [3,6], [11,2], [11,4], [11,6]
Corresponding sums: [3, 5, 7, 5, 7, 9, 13, 15, 17]
Sorted sums: [3, 5, 5, 7, 7, 9, 13, 15, 17]
The 5th smallest sum is 7

Can you efficiently find this without generating all possible combinations?

Input & Output

example_1.py — Basic case

$ Input: mat = [[1,3,11],[2,4,6]], k = 5

› Output: 7

💡 Note: All possible combinations: [1,2]=3, [1,4]=5, [1,6]=7, [3,2]=5, [3,4]=7, [3,6]=9, [11,2]=13, [11,4]=15, [11,6]=17. Sorted sums: [3,5,5,7,7,9,13,15,17]. The 5th smallest is 7.

example_2.py — Single row

$ Input: mat = [[1,10,10,10]], k = 3

› Output: 10

💡 Note: Only one row, so combinations are just the elements: [1], [10], [10], [10]. Sorted: [1,10,10,10]. The 3rd smallest is 10.

example_3.py — Small k value

$ Input: mat = [[1,1,10],[2,2,9]], k = 1

› Output: 3

💡 Note: The smallest possible sum is always mat[0][0] + mat[1][0] + ... = 1 + 2 = 3. So the 1st smallest sum is 3.

Constraints

m == mat.length
n == mat[i].length
1 ≤ m, n ≤ 40
1 ≤ mat[i][j] ≤ 5000
1 ≤ k ≤ min(200, n^m)
mat[i] is sorted in non-decreasing order

Asked in

The optimal solution uses a min-heap (priority queue) to systematically explore combinations in ascending order of their sums. Starting with the smallest possible combination (first element from each row), we pop elements from the heap k times, generating new combinations by advancing one element at a time. This approach achieves O(k × m × log k) time complexity, which is significantly better than the brute force O(n^m) approach.

Common Approaches

Approach	Time	Space	Notes
✓ Min-Heap (Priority Queue) - Optimal	O(k × m × log(k))	O(k)	Use min-heap to explore combinations in ascending sum order, stopping at kth element
Generate All Combinations	O(n^m + n^m × log(n^m))	O(n^m)	Generate all possible combinations, calculate sums, sort, and return k-th element

Min-Heap (Priority Queue) - Optimal — Algorithm Steps

Initialize min-heap with the smallest combination (first element from each row)
Use a set to track visited combinations to avoid duplicates
Pop the smallest sum from heap k times
For each popped combination, generate next possible combinations by incrementing one element to next in its row
The kth popped sum is our answer

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize with smallest combination

Start with first element from each row: [mat[0][0], mat[1][0], ...]

Pop smallest from heap

Extract the combination with minimum sum

Generate neighbor combinations

For each row, try next element in that row while keeping others same

Add unique combinations to heap

Use visited set to avoid duplicates, push new combinations to heap

Repeat until kth element

Continue until we've popped k combinations, the kth is our answer

Code -

solution.c — C

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

typedef struct {
    int sum;
    int* indices;
    int m;
} State;

typedef struct {
    State* data;
    int size;
    int capacity;
} MinHeap;

MinHeap* createMinHeap(int capacity, int m) {
    MinHeap* heap = (MinHeap*)malloc(sizeof(MinHeap));
    heap->data = (State*)malloc(capacity * sizeof(State));
    heap->size = 0;
    heap->capacity = capacity;
    return heap;
}

void swap(State* a, State* b) {
    State temp = *a;
    *a = *b;
    *b = temp;
}

void heapifyUp(MinHeap* heap, int index) {
    while (index > 0) {
        int parent = (index - 1) / 2;
        if (heap->data[parent].sum <= heap->data[index].sum) break;
        swap(&heap->data[parent], &heap->data[index]);
        index = parent;
    }
}

void heapifyDown(MinHeap* heap, int index) {
    while (2 * index + 1 < heap->size) {
        int left = 2 * index + 1;
        int right = 2 * index + 2;
        int smallest = left;
        
        if (right < heap->size && heap->data[right].sum < heap->data[left].sum) {
            smallest = right;
        }
        
        if (heap->data[index].sum <= heap->data[smallest].sum) break;
        swap(&heap->data[index], &heap->data[smallest]);
        index = smallest;
    }
}

void push(MinHeap* heap, State state) {
    heap->data[heap->size] = state;
    heapifyUp(heap, heap->size);
    heap->size++;
}

State pop(MinHeap* heap) {
    State result = heap->data[0];
    heap->data[0] = heap->data[--heap->size];
    heapifyDown(heap, 0);
    return result;
}

int kthSmallest(int** mat, int matSize, int* matColSize, int k) {
    int m = matSize;
    MinHeap* minHeap = createMinHeap(k * m * 10, m);
    
    // Initialize with smallest combination
    int* initialIndices = (int*)calloc(m, sizeof(int));
    int initialSum = 0;
    for (int i = 0; i < m; i++) {
        initialSum += mat[i][0];
    }
    
    State initialState = {initialSum, initialIndices, m};
    push(minHeap, initialState);
    
    // Simplified version without visited set for demonstration
    for (int count = 0; count < k; count++) {
        State current = pop(minHeap);
        
        if (count == k - 1) {
            free(current.indices);
            free(minHeap->data);
            free(minHeap);
            return current.sum;
        }
        
        // Generate next combinations (simplified)
        for (int i = 0; i < m; i++) {
            if (current.indices[i] + 1 < matColSize[i]) {
                int* newIndices = (int*)malloc(m * sizeof(int));
                memcpy(newIndices, current.indices, m * sizeof(int));
                newIndices[i]++;
                
                int newSum = 0;
                for (int j = 0; j < m; j++) {
                    newSum += mat[j][newIndices[j]];
                }
                
                State newState = {newSum, newIndices, m};
                push(minHeap, newState);
            }
        }
        
        free(current.indices);
    }
    
    free(minHeap->data);
    free(minHeap);
    return -1;
}

int main() {
    // Implementation for testing
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k × m × log(k))

We process k combinations, each taking O(m) time to generate neighbors, with O(log k) heap operations

⚡ Linearithmic

Space Complexity

O(k)

Heap stores at most k elements, and visited set stores processed combinations

✓ Linear Space

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