Optimal Partition of String

Optimal Partition of String - Problem

Given a string s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once.

Return the minimum number of substrings in such a partition.

Note that each character should belong to exactly one substring in a partition.

Input & Output

Example 1 — Basic Case

$ Input: s = "abcabc"

› Output: 2

💡 Note: Split into "abc" and "abc". Each substring has unique characters. We cannot do better than 2 partitions because 'a' appears again at position 3.

Example 2 — All Unique

$ Input: s = "abcd"

› Output: 1

💡 Note: All characters are unique, so we only need 1 partition containing the entire string.

Example 3 — All Same

$ Input: s = "aaaa"

› Output: 4

💡 Note: Every character is the same, so each must be in its own partition: "a", "a", "a", "a".

Constraints

1 ≤ s.length ≤ 1000
s consists of only English lowercase letters

Visualization

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Asked in

a Amazon 35 G Google 25 M Microsoft 20

The key insight is to use a greedy approach: scan left to right and start a new partition whenever we encounter a duplicate character. This minimizes partitions because we make each partition as long as possible. Best approach is Greedy with Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Try All Partitions

⏱️ Time: O(2ⁿ × n) Space: O(n)

Try every possible way to partition the string. For each partition, check if all substrings have unique characters, then find the minimum count among valid partitions.

Greedy Approach

⏱️ Time: O(n) Space: O(1)

Use a greedy strategy: scan the string from left to right, keeping track of seen characters. When we encounter a duplicate, start a new partition. This gives the optimal solution because making each partition as long as possible minimizes the total count.

Brute Force - Try All Partitions — Algorithm Steps

Generate all possible partitions of the string
For each partition, verify all substrings have unique characters
Track the minimum partition count among valid partitions

Visualization

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Step-by-Step Walkthrough

Generate Partitions

Create all possible ways to split the string

Validate Each

Check if each partition has unique characters

Find Minimum

Return the partition with minimum count

Code -

solution.c — C

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

bool hasUniqueChars(const char* substring) {
    bool seen[256] = {false};
    for (int i = 0; substring[i]; i++) {
        if (seen[(unsigned char)substring[i]]) return false;
        seen[(unsigned char)substring[i]] = true;
    }
    return true;
}

int backtrack(const char* s, int index, char partitions[][1001], int partitionCount, int len) {
    if (index == len) {
        for (int i = 0; i < partitionCount; i++) {
            if (!hasUniqueChars(partitions[i])) return 1000000;
        }
        return partitionCount;
    }
    
    int minCount = 1000000;
    // Try all possible end positions for current substring
    for (int end = index + 1; end <= len; end++) {
        strncpy(partitions[partitionCount], s + index, end - index);
        partitions[partitionCount][end - index] = '\0';
        
        int count = backtrack(s, end, partitions, partitionCount + 1, len);
        if (count < minCount) minCount = count;
    }
    
    return minCount;
}

int solution(const char* s) {
    char partitions[1001][1001];
    return backtrack(s, 0, partitions, 0, strlen(s));
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2ⁿ × n)

2ⁿ possible partitions, each takes O(n) time to validate

✓ Linear Growth

Space Complexity

O(n)

Recursion stack depth and temporary storage for partitions

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Partitions — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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