703864. Maximum bitonic subarray sum
Maximum Bitonic Subarray Sum
Summary
Given a binary array (0s and 1s), find the maximum sum of bitonic subarrays. A bitonic sequence is either strictly increasing or strictly decreasing.
The problem involves arrays, which is one of the tags mentioned. We'll explore different approaches to solve this problem efficiently.
Detailed Explanation
To approach this problem, we can start by understanding what a bitonic sequence is. A bitonic sequence is either strictly increasing or strictly decreasing. So, if we find all the bitonic sequences in the given array and calculate their sums, we can then return the maximum sum among them.
Let's break down the steps to solve this problem:
- Initialize two variables:
maxSumto store the maximum sum found so far andcurrentSumto keep track of the current sum being calculated. - Iterate over the array from left to right:
- If the current element is 0, it means we're entering a decreasing part of the bitonic sequence. In this case, update
currentSumwith the value of the previous element (since the sequence starts increasing again after this). - If the current element is 1, it means we're in an increasing part of the bitonic sequence. Update
currentSumby adding the value of the current element.
- If the current element is 0, it means we're entering a decreasing part of the bitonic sequence. In this case, update
- Keep updating
maxSumwith the maximum ofmaxSumandcurrentSum. - Finally, return
maxSum.
Time complexity: O(n), where n is the size of the array.
Space complexity: O(1), as we're only using a constant amount of space to store maxSum, currentSum, and some temporary variables.
Optimized Solutions
Java
public int maxBitonicSubarraySum(int[] arr) {
int n = arr.length;
int maxSum = 0, currentSum = 0;
for (int i = 0; i < n; i++) {
if (arr[i] == 0) {
currentSum = currentSum > 0 ? 1 : 0;
} else {
currentSum += arr[i];
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}