LeetCode 1562: Find Latest Group of Size M Solution

Master LeetCode problem 1562 (Find Latest Group of Size M), 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.

1562. Find Latest Group of Size M

Medium

Array Hash Table Binary Search Simulation

Problem Explanation

Explanation

To solve this problem, we can simulate the process of setting bits in the binary string based on the given array arr. We can maintain a set to keep track of the groups of ones formed at each step. For each step, we update the binary string and check if there exists a group of ones of length m. If such a group exists, we update the latest step where this group was found.

Here is the step-by-step algorithm:

Initialize a binary string of size n with all zeros and a set to store the groups of ones.
Iterate over the array arr and update the binary string by setting the bit at position arr[i] to 1.
Update the set of groups by iterating over the binary string and identifying the groups of ones.
Check if there exists a group of ones of length m in the set. If yes, update the latest step where this group was found.
Return the latest step where a group of ones of length m was found or -1 if no such group exists.

The time complexity of this algorithm is O(n) where n is the length of the input array arr. The space complexity is also O(n) to store the binary string and the set of groups.

Solution Code

class Solution {
    public int findLatestStep(int[] arr, int m) {
        int n = arr.length;
        int[] binaryString = new int[n];
        Set<Integer> groups = new HashSet<>();
        int result = -1;
        
        for (int i = 0; i < n; i++) {
            binaryString[arr[i] - 1] = 1;
            groups.clear();
            int count = 0;
            
            for (int j = 0; j < n; j++) {
                if (binaryString[j] == 1) {
                    count++;
                    if (j == 0 || binaryString[j - 1] == 0) {
                        groups.add(count);
                    }
                } else {
                    count = 0;
                }
            }
            
            for (int groupSize : groups) {
                if (groupSize == m) {
                    result = i + 1;
                }
            }
        }
        
        return result;
    }
}

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