Convert an Array Into a 2D Array With Conditions

Convert an Array Into a 2D Array With Conditions - Problem

You are given an integer array nums. You need to create a 2D array from nums satisfying the following conditions:

The 2D array should contain only the elements of the array nums.
Each row in the 2D array contains distinct integers.
The number of rows in the 2D array should be minimal.

Return the resulting array. If there are multiple answers, return any of them.

Note that the 2D array can have a different number of elements on each row.

Input & Output

Example 1 — Basic Case

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

› Output: [[1,3,4,2],[1,3],[1]]

💡 Note: Number 1 appears 3 times (most frequent), so we need 3 rows minimum. Distribute: Row 1 gets [1,3,4,2], Row 2 gets [1,3], Row 3 gets [1]

Example 2 — All Unique

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

› Output: [[1,2,3,4]]

💡 Note: All numbers are unique, so only 1 row needed containing all elements

Example 3 — All Same

$ Input: nums = [1,1,1,1]

› Output: [[1],[1],[1],[1]]

💡 Note: All numbers are the same, so each must go in a separate row to maintain distinctness

Constraints

1 ≤ nums.length ≤ 200
1 ≤ nums[i] ≤ nums.length

Visualization

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Understanding the Visualization

Input Array

Original 1D array with possible duplicates

Group by Frequency

Determine minimum rows needed = max frequency

Distribute Optimally

Create 2D array where each row has distinct elements

Key Takeaway

🎯 Key Insight: The minimum rows needed equals the maximum frequency of any element

Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is that the minimum number of rows equals the maximum frequency of any element. The frequency counting approach is optimal: count each element's frequency, create that many rows, then distribute elements round-robin style. Time: O(n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Frequency Count Optimization	O(n)	O(n)	Count frequencies first, then distribute elements optimally across rows
Brute Force - Try All Combinations	O(n²)	O(n)	Try placing each element in every possible row until finding valid arrangement

Frequency Count Optimization — Algorithm Steps

Count frequency of each number
Create rows equal to max frequency
Distribute each number across different rows
Return the resulting 2D array

Visualization

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Step-by-Step Walkthrough

Count Frequencies

Count how many times each number appears

Create Optimal Rows

Create exactly max_frequency rows

Distribute Round-Robin

Place each number in different rows

Code -

solution.c — C

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

int** solution(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {
    // Simple frequency counting using array (assuming nums are small)
    int freq[1001] = {0}; // assuming nums[i] <= 1000
    int maxFreq = 0;
    
    // Count frequencies
    for (int i = 0; i < numsSize; i++) {
        freq[nums[i]]++;
        if (freq[nums[i]] > maxFreq) {
            maxFreq = freq[nums[i]];
        }
    }
    
    // Create rows
    int** result = (int**)malloc(maxFreq * sizeof(int*));
    *returnColumnSizes = (int*)calloc(maxFreq, sizeof(int));
    *returnSize = maxFreq;
    
    for (int i = 0; i < maxFreq; i++) {
        result[i] = (int*)malloc(1001 * sizeof(int));
    }
    
    // Distribute elements
    for (int num = 0; num <= 1000; num++) {
        if (freq[num] > 0) {
            for (int i = 0; i < freq[num]; i++) {
                result[i][(*returnColumnSizes)[i]] = num;
                (*returnColumnSizes)[i]++;
            }
        }
    }
    
    return result;
}

int main() {
    // Simple test case parsing for demo
    int nums[] = {1, 3, 4, 1, 2, 3, 1};
    int numsSize = 7;
    int returnSize;
    int* returnColumnSizes;
    
    int** result = solution(nums, numsSize, &returnSize, &returnColumnSizes);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("[");
        for (int j = 0; j < returnColumnSizes[i]; j++) {
            printf("%d", result[i][j]);
            if (j < returnColumnSizes[i] - 1) printf(",");
        }
        printf("]");
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

One pass to count frequencies, one pass to distribute elements

✓ Linear Growth

Space Complexity

O(n)

Hash map for frequencies plus result array

⚡ Linearithmic Space

12.5K Views

Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Frequency Count Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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