Split Message Based on Limit

Split Message Based on Limit - Problem

You are given a string message and a positive integer limit.

You must split message into one or more parts based on limit. Each resulting part should have the suffix "<a/b>", where "b" is to be replaced with the total number of parts and "a" is to be replaced with the index of the part, starting from 1 and going up to b. Additionally, the length of each resulting part (including its suffix) should be equal to limit, except for the last part whose length can be at most limit.

The resulting parts should be formed such that when their suffixes are removed and they are all concatenated in order, they should be equal to message. Also, the result should contain as few parts as possible.

Return the parts message would be split into as an array of strings. If it is impossible to split message as required, return an empty array.

Input & Output

Example 1 — Basic Split

$ Input: message = "this is really a long text", limit = 16

› Output: ["this is rea<1/3>", "lly a long <2/3>", "text<3/3>"]

💡 Note: Split into 3 parts: Part 1 has 11 chars + 5 char suffix = 16 total. Part 2 has 11 chars + 5 char suffix = 16 total. Part 3 has 4 chars + 5 char suffix = 9 total (≤ 16).

Example 2 — Impossible Split

$ Input: message = "short", limit = 15

› Output: ["short<1/1>"]

💡 Note: Single part: "short" (5 chars) + "<1/1>" (5 chars) = 10 total ≤ 15, so this is possible.

Example 3 — Edge Case Single Part

$ Input: message = "ab", limit = 8

› Output: ["ab<1/1>"]

💡 Note: Single part: "ab" (2 chars) + "<1/1>" (5 chars) = 7 chars ≤ 8 limit.

Constraints

1 ≤ message.length ≤ 10⁴
1 ≤ limit ≤ 10⁴

Visualization

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Asked in

G Google 25 f Meta 18 a Amazon 15 M Microsoft 12

The key insight is that the suffix length depends on the total number of parts, so we must try different part counts to find the minimum valid split. The optimal approach uses enumeration or binary search to find the minimum number of parts, then validates if the message can be split accordingly. Time: O(b × n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Binary Search on Answer	O(log n × n)	O(n)	Use binary search to find the minimum number of parts needed
Brute Force Enumeration	O(b × n)	O(n)	Try every possible number of parts from 1 upward until we find a valid split

Binary Search on Answer — Algorithm Steps

Step 1: Set search range from 1 to message.length parts
Step 2: For mid value, check if message can be split into mid parts
Step 3: If possible, search left half; otherwise search right half
Step 4: Build and return the actual split for the minimum valid parts

Visualization

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Step-by-Step Walkthrough

Search Range

Set left=1, right=message.length

Test Middle

Check if mid parts can accommodate the message

Narrow Range

Adjust search range based on feasibility

Code -

solution.c — C

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

bool canSplit(char* message, int limit, int parts) {
    // Check if we can split message into 'parts' number of parts
    char sampleSuffix[50];
    sprintf(sampleSuffix, "<%d/%d>", parts, parts);
    int suffixLen = strlen(sampleSuffix);
    if (suffixLen >= limit) {
        return false;
    }
    
    int availableSpace = limit - suffixLen;
    long long totalAvailable = (long long)availableSpace * (parts - 1) + (limit - suffixLen);
    return totalAvailable >= strlen(message);
}

char** buildSplit(char* message, int limit, int parts, int* returnSize) {
    // Actually build the split with 'parts' number of parts
    int n = strlen(message);
    char sampleSuffix[50];
    sprintf(sampleSuffix, "<%d/%d>", parts, parts);
    int suffixLen = strlen(sampleSuffix);
    int availableSpace = limit - suffixLen;
    
    char** result = (char**)malloc(parts * sizeof(char*));
    int pos = 0;
    *returnSize = 0;
    
    for (int i = 1; i <= parts; i++) {
        char suffix[50];
        sprintf(suffix, "<%d/%d>", i, parts);
        
        result[i-1] = (char*)malloc((limit + 1) * sizeof(char));
        
        if (i == parts) { // Last part
            int remainingLen = n - pos;
            if (remainingLen + strlen(suffix) > limit) {
                // Free allocated memory
                for (int j = 0; j < i; j++) {
                    free(result[j]);
                }
                free(result);
                return NULL;
            }
            strncpy(result[i-1], message + pos, remainingLen);
            result[i-1][remainingLen] = '\0';
            strcat(result[i-1], suffix);
            (*returnSize)++;
        } else {
            int charsToTake = (availableSpace < n - pos) ? availableSpace : n - pos;
            strncpy(result[i-1], message + pos, charsToTake);
            result[i-1][charsToTake] = '\0';
            strcat(result[i-1], suffix);
            pos += charsToTake;
            (*returnSize)++;
        }
    }
    
    if (pos == n) {
        return result;
    }
    
    // Free memory if failed
    for (int j = 0; j < *returnSize; j++) {
        free(result[j]);
    }
    free(result);
    *returnSize = 0;
    return NULL;
}

char** solution(char* message, int limit, int* returnSize) {
    int n = strlen(message);
    *returnSize = 0;
    
    // Binary search for minimum number of parts
    int left = 1, right = n + 1;
    int minParts = -1;
    
    while (left <= right) {
        int mid = (left + right) / 2;
        if (canSplit(message, limit, mid)) {
            minParts = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    
    if (minParts == -1) {
        return NULL;
    }
    
    return buildSplit(message, limit, minParts, returnSize);
}

int main() {
    char message[1000];
    fgets(message, sizeof(message), stdin);
    message[strcspn(message, "\n")] = 0; // Remove newline
    
    int limit;
    scanf("%d", &limit);
    
    int returnSize;
    char** result = solution(message, limit, &returnSize);
    
    // Format output as JSON array
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printf("\"%s\"", result[i]);
        free(result[i]);
    }
    printf("]\n");
    
    if (result) free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n × n)

Binary search O(log n) iterations, each requiring O(n) validation

⚡ Linearithmic

Space Complexity

O(n)

Store the result array containing message characters

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search on Answer — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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