Find the Shortest Superstring

Find the Shortest Superstring - Problem

Find the Shortest Superstring is a fascinating string optimization problem that challenges you to find the most efficient way to combine multiple strings into one compact superstring.

Given an array of strings words, your task is to construct the shortest possible string that contains each string in words as a substring. Think of it as creating a master string that encompasses all input strings with maximum overlap.

Key Points:
• If multiple valid strings exist with the same minimum length, return any of them
• No string in the input is a substring of another string
• The goal is to maximize overlaps between strings to minimize total length

Example: For ["catg", "ctaagt", "gcta"], one possible superstring is "gctaagttcatg" which contains all three strings with optimal overlapping.

Input & Output

example_1.py — Basic Case

$ Input: ["catg", "ctaagt", "gcta"]

› Output: "gctaagttcatg"

💡 Note: The strings can be arranged as gcta → catg → ctaagt with overlaps: gcta+catg overlap by 'cat' (3 chars), catg+ctaagt overlap by 'ct' (2 chars). Result: gcta + g + aagt = gctaagttcatg

example_2.py — Simple Chain

$ Input: ["ab", "bc", "cd"]

› Output: "abcd"

💡 Note: Perfect chain with single character overlaps: ab → bc → cd becomes ab + c + d = abcd

example_3.py — No Overlaps

$ Input: ["abc", "def", "ghi"]

› Output: "abcdefghi"

💡 Note: When no overlaps exist between any strings, we simply concatenate them in any order. Total length equals sum of all string lengths.

Constraints

1 ≤ words.length ≤ 12
1 ≤ words[i].length ≤ 20
words[i] consists of lowercase English letters
No string is a substring of another string
The answer is guaranteed to be unique

Visualization

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Asked in

G Google 38 a Amazon 24 f Meta 19 ⊞ Microsoft 15

The optimal solution uses Dynamic Programming with Bitmasks to model this as a Traveling Salesman Problem variant. We precompute all pairwise overlaps, then use DP states dp[mask][i] to track the minimum superstring length using strings in 'mask' and ending at string 'i'. Time complexity: O(n² × 2ⁿ), which is efficient for the constraint n ≤ 12.

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Brute Force (Try All Permutations)

⏱️ Time: O(n! × n²) Space: O(n)

This approach tries every possible ordering of the input strings and constructs a superstring for each permutation by greedily overlapping adjacent strings. It then returns the shortest superstring found.

Dynamic Programming with Bitmask

⏱️ Time: O(n² × 2ⁿ) Space: O(n × 2ⁿ)

This approach models the problem as a variant of the Traveling Salesman Problem. We use dynamic programming with bitmasks to track which strings have been used, and for each state, we track the minimum cost to reach that state ending at each possible string.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

typedef struct Element {
    int isArray;
    int value;
    struct Element* array;
    int arraySize;
} Element;

typedef struct StackItem {
    Element* element;
    int depth;
} StackItem;

Element* createInt(int val) {
    Element* elem = malloc(sizeof(Element));
    elem->isArray = 0;
    elem->value = val;
    elem->array = NULL;
    elem->arraySize = 0;
    return elem;
}

Element* createArray(Element** elements, int size) {
    Element* elem = malloc(sizeof(Element));
    elem->isArray = 1;
    elem->value = 0;
    elem->array = malloc(size * sizeof(Element));
    elem->arraySize = size;
    for (int i = 0; i < size; i++) {
        elem->array[i] = *elements[i];
    }
    return elem;
}

Element** solution(Element* arr, int arrSize, int n, int* returnSize) {
    StackItem* stack = malloc(1000 * sizeof(StackItem));
    Element** result = malloc(1000 * sizeof(Element*));
    int stackSize = 0;
    *returnSize = 0;
    
    // Push elements in reverse order
    for (int i = arrSize - 1; i >= 0; i--) {
        stack[stackSize].element = &arr[i];
        stack[stackSize].depth = 0;
        stackSize++;
    }
    
    while (stackSize > 0) {
        stackSize--;
        StackItem item = stack[stackSize];
        
        if (item.element->isArray && item.depth < n) {
            for (int i = item.element->arraySize - 1; i >= 0; i--) {
                stack[stackSize].element = &item.element->array[i];
                stack[stackSize].depth = item.depth + 1;
                stackSize++;
            }
        } else {
            result[(*returnSize)++] = item.element;
        }
    }
    
    free(stack);
    return result;
}

void printElement(Element* elem) {
    if (!elem->isArray) {
        printf("%d", elem->value);
    } else {
        printf("[");
        for (int i = 0; i < elem->arraySize; i++) {
            if (i > 0) printf(",");
            printElement(&elem->array[i]);
        }
        printf("]");
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple hardcoded parsing for test cases
    Element* arr = malloc(10 * sizeof(Element));
    int arrSize = 0;
    
    if (strstr(line, "[1,2,[3,4,[5,6]]]")) {
        arr[0] = *createInt(1);
        arr[1] = *createInt(2);
        Element* sub = malloc(3 * sizeof(Element));
        sub[0] = *createInt(3);
        sub[1] = *createInt(4);
        Element* subsub = malloc(2 * sizeof(Element));
        subsub[0] = *createInt(5);
        subsub[1] = *createInt(6);
        sub[2].isArray = 1;
        sub[2].array = subsub;
        sub[2].arraySize = 2;
        arr[2].isArray = 1;
        arr[2].array = sub;
        arr[2].arraySize = 3;
        arrSize = 3;
    }
    
    int n;
    scanf("%d", &n);
    
    int returnSize;
    Element** result = solution(arr, arrSize, n, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printElement(result[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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