LeetCode 2963: Count the Number of Good Partitions Solution

Master LeetCode problem 2963 (Count the Number of Good Partitions), 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.

2963. Count the Number of Good Partitions

Hard

Array Hash Table Math Combinatorics

Problem Explanation

Explanation

To solve this problem, we can iterate through the array from left to right and keep track of the last index where each number occurred. We can use a hashmap to store this information. Then, for each index in the array, we can calculate the number of good partitions that end at that index by considering all possible partitions that end with the current number. Finally, we sum up the total number of good partitions over all indices.

Algorithm:

Initialize a hashmap to store the last index of each number.
Initialize a variable to keep track of the total number of good partitions.
Iterate through the array from left to right.
For each index, calculate the number of good partitions that end at that index by considering all possible partitions that end with the current number.
Update the total number of good partitions.
Return the total number of good partitions modulo 10^9 + 7.

Time Complexity: O(n), where n is the length of the input array. Space Complexity: O(n), for the hashmap to store the last index of each number.

Solution Code

class Solution {
    public int countGoodPartitions(int[] nums) {
        long mod = 1000000007;
        Map<Integer, Integer> lastIndices = new HashMap<>();
        long goodPartitions = 0;
        
        for (int i = 0; i < nums.length; i++) {
            int num = nums[i];
            int lastIndex = lastIndices.getOrDefault(num, -1);
            int partitions = i - lastIndex;
            goodPartitions += partitions;
            lastIndices.put(num, i);
        }
        
        return (int)(goodPartitions % mod);
    }
}

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