Reshape Data: Pivot - Practice Coding Problems

Reshape Data: Pivot - Problem

Databse Easy

You have a DataFrame weather with the following structure:

Column Name	Type
city	object
month	object
temperature	int

Write a solution to pivot the data so that each row represents temperatures for a specific month, and each city is a separate column.

The result should transform data from a long format (multiple rows per month) to a wide format (one row per month with cities as columns).

Input & Output

Example 1 — Basic Weather Data

$ Input: weather_data = [["New York", "January", 25], ["Los Angeles", "January", 75], ["New York", "February", 30], ["Los Angeles", "February", 80]]

› Output: [["month", "Los Angeles", "New York"], ["February", 80, 30], ["January", 75, 25]]

💡 Note: Transform from long format (city-month-temp rows) to wide format where months become rows and cities become columns. Cities are sorted alphabetically.

Example 2 — Missing Data

$ Input: weather_data = [["Miami", "March", 85], ["Seattle", "April", 55], ["Miami", "April", 88]]

› Output: [["month", "Miami", "Seattle"], ["April", 88, 55], ["March", 85, null]]

💡 Note: When a city has no temperature data for a specific month, the pivot table shows null for that cell (Seattle has no March data).

Example 3 — Single Record

$ Input: weather_data = [["Chicago", "December", 15]]

› Output: [["month", "Chicago"], ["December", 15]]

💡 Note: Even with minimal data, the pivot structure is maintained with proper headers and single data row.

Constraints

1 ≤ weather_data.length ≤ 1000
weather_data[i].length = 3
weather_data[i] = [city, month, temperature]
All city and month names are non-empty strings
All temperatures are integers

Visualization

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Understanding the Visualization

Input Data

Weather records in long format: [city, month, temperature]

Pivot Transform

Reshape data with months as rows and cities as columns

Output Table

Structured table format with proper headers

Key Takeaway

🎯 Key Insight: Pivoting transforms analytical perspective by making cities comparable across months in a structured table format

Asked in

f Meta 15 G Google 12 ⊞ Microsoft 10 a Amazon 8

The key insight is to transform row-wise data into column-wise by using cities as column headers and months as row indices. The optimal approach uses a single-pass algorithm with hash maps to build the pivot structure efficiently. Time: O(n + m log m + c log c), Space: O(m×c).

Common Approaches

Approach	Time	Space	Notes
✓ Pandas-like Pivot Implementation	O(n + m log m + c log c)	O(m×c)	Efficient single-pass pivot using optimized data structures
Manual Dictionary Building	O(n)	O(m×c)	Manually build nested dictionaries to group data by month and city

Pandas-like Pivot Implementation — Algorithm Steps

Step 1: Single pass to collect unique values and build pivot map
Step 2: Sort dimensions for consistent output
Step 3: Generate result matrix efficiently

Visualization

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Step-by-Step Walkthrough

Single Pass

Process all records once, building pivot map and collecting unique values

Sort Dimensions

Sort cities and months for consistent output ordering

Generate Matrix

Build final result matrix efficiently using map lookups

Code -

solution.c — C

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

#define MAX_RECORDS 100
#define MAX_CITIES 20
#define MAX_MONTHS 20
#define MAX_STR_LEN 50

typedef struct {
    char cities[MAX_CITIES][MAX_STR_LEN];
    char months[MAX_MONTHS][MAX_STR_LEN];
    int temps[MAX_MONTHS][MAX_CITIES];
    int hasTempo[MAX_MONTHS][MAX_CITIES];
    int cityCount;
    int monthCount;
} PivotData;

int findOrAddString(char arr[][MAX_STR_LEN], int* count, const char* str, int maxCount) {
    for (int i = 0; i < *count; i++) {
        if (strcmp(arr[i], str) == 0) {
            return i;
        }
    }
    if (*count < maxCount) {
        strcpy(arr[*count], str);
        return (*count)++;
    }
    return -1;
}

PivotData solution(char weatherData[][3][MAX_STR_LEN], int dataSize) {
    PivotData result = {0};
    
    // Build pivot structure
    for (int i = 0; i < dataSize; i++) {
        int cityIdx = findOrAddString(result.cities, &result.cityCount, weatherData[i][0], MAX_CITIES);
        int monthIdx = findOrAddString(result.months, &result.monthCount, weatherData[i][1], MAX_MONTHS);
        
        if (cityIdx >= 0 && monthIdx >= 0) {
            result.temps[monthIdx][cityIdx] = atoi(weatherData[i][2]);
            result.hasTempo[monthIdx][cityIdx] = 1;
        }
    }
    
    return result;
}

int main() {
    char weatherData[MAX_RECORDS][3][MAX_STR_LEN];
    int dataSize = 0;
    
    PivotData result = solution(weatherData, dataSize);
    printf("[]");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m log m + c log c)

O(n) to process records, O(m log m) to sort months, O(c log c) to sort cities

⚡ Linearithmic

Space Complexity

O(m×c)

Store pivot table with m months × c cities plus temporary sets

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Pandas-like Pivot Implementation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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