Construct Smallest Number From DI String

Construct Smallest Number From DI String - Problem

You are given a 0-indexed string pattern of length n consisting of the characters 'I' meaning increasing and 'D' meaning decreasing.

A 0-indexed string num of length n + 1 is created using the following conditions:

num consists of the digits '1' to '9', where each digit is used at most once.
If pattern[i] == 'I', then num[i] < num[i + 1].
If pattern[i] == 'D', then num[i] > num[i + 1].

Return the lexicographically smallest possible string num that meets the conditions.

Input & Output

Example 1 — Basic Decreasing-Increasing

$ Input: pattern = "DI"

› Output: "213"

💡 Note: For pattern "DI", we need num[0] > num[1] (D) and num[1] < num[2] (I). Using digits [1,2,3], the arrangement "213" satisfies: 2 > 1 (D) and 1 < 3 (I).

Example 2 — All Increasing

$ Input: pattern = "III"

› Output: "1234"

💡 Note: Pattern "III" means all positions should be increasing. The lexicographically smallest arrangement is "1234": 1 < 2 < 3 < 4.

Example 3 — All Decreasing

$ Input: pattern = "DDI"

› Output: "3214"

💡 Note: Pattern "DDI" needs two decreasing positions followed by increasing. Starting with [1,2,3,4], reverse the D-segment [1,2,3] to get [3,2,1,4], giving "3214".

Constraints

1 ≤ pattern.length ≤ 8
pattern contains only characters 'I' and 'D'

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 12 F Facebook 8

The key insight is to use a stack to handle decreasing sequences efficiently. When we encounter 'D', we accumulate digits in a stack, and when we see 'I' (or reach the end), we pop all digits to reverse the decreasing segment. Best approach is Stack Method with Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force - Try All Permutations

⏱️ Time: O((n+1)!) Space: O(n)

Try every permutation of digits 1 to n+1, check if each follows the I/D pattern, and return the lexicographically smallest valid permutation.

Greedy Approach

⏱️ Time: O(n) Space: O(n)

Start with consecutive integers 1 to n+1, then use a greedy strategy to arrange them by reversing segments where we have consecutive D's to get the lexicographically smallest result.

Stack Approach

⏱️ Time: O(n) Space: O(n)

Build the result by placing digits 1 to n+1 in sequence, using a stack to reverse decreasing subsequences when encountering 'I' or at the end.

Brute Force - Try All Permutations — Algorithm Steps

Generate all permutations of digits 1 to n+1
For each permutation, check if it satisfies the I/D pattern
Return the lexicographically smallest valid permutation

Visualization

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Step-by-Step Walkthrough

Generate Permutations

Create all possible arrangements of digits 1 to n+1

Check Pattern

For each permutation, verify it follows I/D rules

Return First Valid

Return the lexicographically smallest valid arrangement

Code -

solution.c — C

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

void swap(int* a, int* b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

int checkValid(int* nums, char* pattern, int n) {
    for (int i = 0; i < n; i++) {
        if (pattern[i] == 'I' && nums[i] >= nums[i + 1]) {
            return 0;
        }
        if (pattern[i] == 'D' && nums[i] <= nums[i + 1]) {
            return 0;
        }
    }
    return 1;
}

char* permute(int* nums, char* pattern, int start, int n, char* result) {
    if (start == n + 1) {
        if (checkValid(nums, pattern, n)) {
            for (int i = 0; i <= n; i++) {
                result[i] = '0' + nums[i];
            }
            result[n + 1] = '\0';
            return result;
        }
        return NULL;
    }
    
    for (int i = start; i <= n; i++) {
        swap(&nums[start], &nums[i]);
        char* res = permute(nums, pattern, start + 1, n, result);
        if (res != NULL) {
            return res;
        }
        swap(&nums[start], &nums[i]);
    }
    return NULL;
}

char* solution(char* pattern) {
    int n = strlen(pattern);
    int* nums = (int*)malloc((n + 1) * sizeof(int));
    char* result = (char*)malloc((n + 2) * sizeof(char));
    
    for (int i = 0; i <= n; i++) {
        nums[i] = i + 1;
    }
    
    char* res = permute(nums, pattern, 0, n, result);
    free(nums);
    
    return res ? res : "";
}

int main() {
    char pattern[10];
    fgets(pattern, sizeof(pattern), stdin);
    pattern[strcspn(pattern, "\n")] = 0;
    
    char* result = solution(pattern);
    printf("%s\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O((n+1)!)

Generate (n+1)! permutations and check each one

✓ Linear Growth

Space Complexity

O(n)

Store current permutation and result string

⚡ Linearithmic Space

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Medium Frequency

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Related Problems

Common Approaches

Brute Force - Try All Permutations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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