LeetCode 1849: Splitting a String Into Descending Consecutive Values Solution

Master LeetCode problem 1849 (Splitting a String Into Descending Consecutive Values), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1849. Splitting a String Into Descending Consecutive Values

Medium

String Backtracking

Problem Explanation

Explanation:

To solve this problem, we can use a recursive approach to check all possible splits of the input string. At each step, we check if the current substring forms a valid descending consecutive sequence. If it does, we recursively check the remaining part of the string. We keep track of the previous number in the descending sequence and the remaining string to process.

Start by iterating through different lengths of the first substring.
For each length, convert the substring to an integer and check if it forms a valid descending sequence.
If it does, recursively check the remaining part of the string starting from the next index with the next expected number in the sequence.
Repeat the above steps until we either reach the end of the string or find an invalid split.

Time Complexity: O(2^n) where n is the length of the input string. In the worst case, we consider all possible splits. Space Complexity: O(n) for the recursion stack.

Solution Code

class Solution {
    public boolean splitString(String s) {
        return backtrack(s, 0, 0);
    }
    
    private boolean backtrack(String s, int index, long prev) {
        if (index == s.length()) {
            return true;
        }
        
        long num = 0;
        for (int i = index; i < s.length(); i++) {
            num = num * 10 + (s.charAt(i) - '0');
            if (num < prev) {
                continue;
            }
            if (prev == 0 || num == prev - 1) {
                if (backtrack(s, i + 1, num)) {
                    return true;
                }
            }
        }
        
        return false;
    }
}

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