LeetCode 1585: Check If String Is Transformable With Substring Sort Operations Solution

Master LeetCode problem 1585 (Check If String Is Transformable With Substring Sort Operations), a hard 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.

1585. Check If String Is Transformable With Substring Sort Operations

Hard

String Greedy Sorting

Problem Explanation

Explanation

To solve this problem, we can iterate over both strings from left to right and keep track of the indices where each digit occurs in both strings. We need to ensure that the relative order of digits in s is the same as in t.

We can achieve this by ensuring that for each digit d, the number of occurrences of d in s before the current index should be less than or equal to the number of occurrences of d in t before the current index.

We can use a data structure like an array or a hashmap to keep track of the indices of each digit in both strings. Then, we iterate over the strings and check if the constraint mentioned above is satisfied.

If at any point, the constraint is violated, then it is not possible to transform s into t, and we return false. Otherwise, we return true at the end.

Time Complexity

The time complexity of this approach is O(n), where n is the length of the strings s and t.

Space Complexity

The space complexity is O(1) because we are using a fixed-size array or hashmap to store the occurrences of digits.

Solution Code

public boolean isTransformable(String s, String t) {
    int n = s.length();
    
    int[] sIndices = new int[10];
    int[] tIndices = new int[10];
    
    for (int i = 0; i < n; i++) {
        int d = s.charAt(i) - '0';
        
        if (sIndices[d] >= tIndices[d]) {
            sIndices[d]++;
        } else {
            for (int j = d - 1; j >= 0; j--) {
                if (sIndices[j] < tIndices[j]) {
                    return false;
                }
            }
            
            tIndices[d]++;
        }
    }
    
    return true;
}

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