LeetCode 1856: Maximum Subarray Min-Product Solution
Master LeetCode problem 1856 (Maximum Subarray Min-Product), 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.
1856. Maximum Subarray Min-Product
Problem Explanation
Explanation
To solve this problem, we can iterate through the array and for each element, we find the left and right boundaries of the subarray with the minimum value equal to the current element. We calculate the sum of elements within this subarray and update the maximum min-product accordingly. This approach ensures that we consider all possible subarrays with the minimum value at the current element.
Time complexity: O(N) where N is the number of elements in the array. Space complexity: O(N) for storing the left and right boundaries.
Solution Code
class Solution {
public int maxSumMinProduct(int[] nums) {
long[] prefixSum = new long[nums.length + 1];
for (int i = 0; i < nums.length; i++) {
prefixSum[i + 1] = prefixSum[i] + nums[i];
}
Stack<Integer> stack = new Stack<>();
long[] left = new long[nums.length];
long[] right = new long[nums.length];
for (int i = 0; i < nums.length; i++) {
while (!stack.isEmpty() && nums[stack.peek()] >= nums[i]) {
stack.pop();
}
left[i] = stack.isEmpty() ? -1 : stack.peek();
stack.push(i);
}
stack.clear();
for (int i = nums.length - 1; i >= 0; i--) {
while (!stack.isEmpty() && nums[stack.peek()] >= nums[i]) {
stack.pop();
}
right[i] = stack.isEmpty() ? nums.length : stack.peek();
stack.push(i);
}
long maxProduct = 0;
for (int i = 0; i < nums.length; i++) {
long sum = prefixSum[(int) right[i]] - prefixSum[(int) left[i] + 1];
maxProduct = Math.max(maxProduct, nums[i] * sum);
}
return (int) (maxProduct % 1000000007);
}
}Try It Yourself
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