LeetCode 1537: Get the Maximum Score Solution

Master LeetCode problem 1537 (Get the Maximum Score), 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.

1537. Get the Maximum Score

Hard

Array Two Pointers Dynamic Programming Greedy

Problem Explanation

Explanation:

To solve this problem, we can iterate through both arrays simultaneously while keeping track of the sum of unique values encountered so far in each array. When we encounter a common value in both arrays, we update our path to switch to the other array to maximize our score. Finally, we return the maximum score obtained modulo 10^9 + 7.

Algorithm:

Initialize pointers i and j to 0 for nums1 and nums2 respectively, and initialize sum1 and sum2 to 0 for tracking cumulative sums.
While both pointers are within bounds:
- If nums1[i] < nums2[j], add nums1[i] to sum1 and increment i.
- If nums1[i] > nums2[j], add nums2[j] to sum2 and increment j.
- If nums1[i] == nums2[j], update the path by taking the maximum sum of sum1 and sum2 plus the common value, and reset sum1 and sum2 to this new maximum sum.
After traversal, update the final score by taking the maximum of sum1 and sum2 and return the score modulo 10^9 + 7.

Time Complexity: O(n + m) where n and m are the lengths of nums1 and nums2 respectively. Space Complexity: O(1)

Solution Code

class Solution {
    public int maxSum(int[] nums1, int[] nums2) {
        long sum1 = 0, sum2 = 0, res = 0;
        int i = 0, j = 0;
        int mod = 1000000007;
        
        while (i < nums1.length || j < nums2.length) {
            if (i < nums1.length && (j == nums2.length || nums1[i] < nums2[j])) {
                sum1 += nums1[i++];
            } else if (j < nums2.length && (i == nums1.length || nums1[i] > nums2[j])) {
                sum2 += nums2[j++];
            } else {
                res += Math.max(sum1, sum2) + nums1[i];
                res %= mod;
                sum1 = 0;
                sum2 = 0;
                i++;
                j++;
            }
        }
        
        res += Math.max(sum1, sum2);
        return (int)(res % mod);
    }
}

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