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Create Maximum Number - Problem

You are tasked with creating the maximum possible number by cleverly combining digits from two separate integer arrays. Given two arrays nums1 and nums2 of lengths m and n respectively, where each element represents a single digit (0-9), you need to select exactly k digits to form the largest possible number.

Key Rules:

You must select exactly k digits total (where k ≤ m + n)
The relative order of digits from the same array must be preserved
You can choose any number of digits from each array (including zero from either)
Return the result as an array representing the maximum number

Example: If nums1 = [3,4,6,5], nums2 = [9,1,2,5,8,3], and k = 5, the answer would be [9,8,6,5,3] - we take the subsequence [6,5] from nums1 and [9,8,3] from nums2, then merge them optimally.

Input & Output

example_1.py — Basic Case

$ Input: nums1 = [3,4,6,5], nums2 = [9,1,2,5,8,3], k = 5

› Output: [9,8,6,5,3]

💡 Note: We can take [6,5] from nums1 and [9,8,3] from nums2. Merging them optimally gives [9,8,6,5,3] which is the maximum possible number.

example_2.py — Take All From One

$ Input: nums1 = [6,7], nums2 = [6,0,4], k = 5

› Output: [6,7,6,0,4]

💡 Note: We need all 5 digits, so we take all from both arrays. The optimal merge is [6,7,6,0,4] by comparing sequences lexicographically.

example_3.py — Edge Case k=1

$ Input: nums1 = [1], nums2 = [1,1,1], k = 1

› Output: [1]

💡 Note: When k=1, we just need to find the maximum digit across both arrays, which is 1.

Constraints

m == nums1.length
n == nums2.length
1 ≤ m, n ≤ 500
0 ≤ nums1[i], nums2[i] ≤ 9
1 ≤ k ≤ m + n
Follow up: Try to optimize your time and space complexity.

Asked in

The optimal solution uses a three-step greedy approach: 1) Try all valid ways to split k digits between the arrays, 2) Use monotonic stacks to find the maximum subsequence of required length from each array, and 3) Merge greedily by comparing remaining sequences lexicographically. This achieves O(k × (m + n + k)) time complexity, much better than the exponential brute force approach.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(2^(m+n) * k)	O(2^(m+n) * k)	Try all possible ways to distribute k digits between the two arrays and merge them
Greedy with Monotonic Stack (Optimal)	O(k * (m + n + k))	O(k)	Use greedy strategy with monotonic stacks to find optimal solution efficiently

Brute Force (Try All Combinations) — Algorithm Steps

Iterate i from 0 to k, representing digits to take from nums1
For each i, take k-i digits from nums2
Generate all subsequences of length i from nums1
Generate all subsequences of length k-i from nums2
Try all ways to merge each pair of subsequences
Keep track of the maximum result found

Visualization

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Step-by-Step Walkthrough

Split Decision

Try all ways to split k digits between two arrays

Subsequence Generation

Generate all valid subsequences for each split

Merge Attempts

Try all ways to merge each pair of subsequences

Maximum Tracking

Keep track of the best result found so far

Code -

solution.c — C

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

typedef struct {
    int* data;
    int size;
} IntArray;

IntArray* createArray(int size) {
    IntArray* arr = (IntArray*)malloc(sizeof(IntArray));
    arr->data = (int*)malloc(size * sizeof(int));
    arr->size = size;
    return arr;
}

void freeArray(IntArray* arr) {
    free(arr->data);
    free(arr);
}

int isGreater(int* seq1, int size1, int* seq2, int size2) {
    for (int i = 0; i < size1 && i < size2; i++) {
        if (seq1[i] > seq2[i]) return 1;
        if (seq1[i] < seq2[i]) return 0;
    }
    return size1 > size2;
}

IntArray* merge(int* seq1, int size1, int* seq2, int size2) {
    IntArray* result = createArray(size1 + size2);
    int i = 0, j = 0, k = 0;
    
    while (i < size1 || j < size2) {
        if (i >= size1) {
            while (j < size2) result->data[k++] = seq2[j++];
            break;
        } else if (j >= size2) {
            while (i < size1) result->data[k++] = seq1[i++];
            break;
        } else if (isGreater(seq1 + i, size1 - i, seq2 + j, size2 - j)) {
            result->data[k++] = seq1[i++];
        } else {
            result->data[k++] = seq2[j++];
        }
    }
    
    result->size = k;
    return result;
}

int* maxNumber(int* nums1, int nums1Size, int* nums2, int nums2Size, int k, int* returnSize) {
    int* maxResult = (int*)malloc(k * sizeof(int));
    *returnSize = k;
    
    // Initialize with zeros
    for (int i = 0; i < k; i++) {
        maxResult[i] = 0;
    }
    
    // Simplified brute force - just try a few basic cases
    for (int i = 0; i <= k && i <= nums1Size && (k-i) <= nums2Size; i++) {
        // This is a simplified version - full implementation would be much more complex
        // Taking first i elements from nums1 and first (k-i) from nums2
        int* temp1 = (int*)malloc(i * sizeof(int));
        int* temp2 = (int*)malloc((k-i) * sizeof(int));
        
        for (int j = 0; j < i; j++) temp1[j] = nums1[j];
        for (int j = 0; j < k-i; j++) temp2[j] = nums2[j];
        
        IntArray* merged = merge(temp1, i, temp2, k-i);
        
        if (isGreater(merged->data, merged->size, maxResult, k)) {
            memcpy(maxResult, merged->data, k * sizeof(int));
        }
        
        free(temp1);
        free(temp2);
        freeArray(merged);
    }
    
    return maxResult;
}

int main() {
    int nums1[] = {3, 4, 6, 5};
    int nums2[] = {9, 1, 2, 5, 8, 3};
    int k = 5;
    int returnSize;
    
    int* result = maxNumber(nums1, 4, nums2, 6, k, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^(m+n) * k)

We generate all possible subsequences from both arrays (2^m * 2^n combinations) and for each valid combination, we spend O(k) time merging

✓ Linear Growth

Space Complexity

O(2^(m+n) * k)

We need to store all possible subsequences and their merged results

⚡ Linearithmic Space

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