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‎MaximumSumCircularSubarray/maximum_sum_circular_subarray.dart

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/*
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-* 918. Maximum Sum Circular SubArray *-
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Given a circular integer array nums of length n, return the maximum possible sum of a non-empty sub-array of nums.
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A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].
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A sub-array may only include each element of the fixed buffer nums at most once. Formally, for a sub-array nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.
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Example 1:
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Input: nums = [1,-2,3,-2]
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Output: 3
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Explanation: Sub-array [3] has maximum sum 3.
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Example 2:
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Input: nums = [5,-3,5]
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Output: 10
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Explanation: Sub-array [5,5] has maximum sum 5 + 5 = 10.
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Example 3:
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Input: nums = [-3,-2,-3]
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Output: -2
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Explanation: Sub-array [-2] has maximum sum -2.
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Constraints:
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n == nums.length
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1 <= n <= 3 * 104
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-3 * 104 <= nums[i] <= 3 * 104
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*/
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import 'dart:math';
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class A {
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int maxSubarraySumCircular(List<int> nums) {
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// max = int_min
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int totalSum = 0, maxSum = -30012, curMax = 0, minSum = 30012, curMin = 0;
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for (int x in nums) {
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curMax = max(x, curMax + x); //update the current max sub-array sum
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maxSum = max(maxSum, curMax); //update the overall max sub-array sum
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curMin = min(x, curMin + x); //update the current min sub-array sum
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minSum = min(minSum, curMin); //update the overall min sub-array sum
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totalSum += x;
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}
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return maxSum > 0 ? max(maxSum, totalSum - minSum) : maxSum;
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}
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}
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class B {
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int kadane(List<int> nums) {
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int local = nums[0];
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int global = nums[0];
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for (int i = 1; i < nums.length; i++) {
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local = max(nums[i], local + nums[i]);
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global = max(global, local);
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}
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return global;
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}
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// case 1: max subarray sum in [0 .. n - 1]
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// i.e. kadane's algo
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// case 2. circular subarray in [0 .. | n - 1 .. | .. 2 * n - 1]
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// i.e. total sum - min subarray sum in [0 .. n - 1]
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int maxSubarraySumCircular(List<int> nums) {
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int n = nums.length;
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// use kadane's algo to find out max sub array sum (case 1)
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int maxSubArraySum = kadane(nums);
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// handle cases like [-3,-2,-3]
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if (maxSubArraySum < 0) return maxSubArraySum;
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// calculate the total sum
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int totalSum = 0;
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// in order to use the same kadane function, we flip the sign
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for (int i = 0; i < n; i++) {
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totalSum += nums[i];
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nums[i] = -nums[i];
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}
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// use kadane's algo to find out min sub array sum
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int minSubArraySum = kadane(nums);
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// compare case 1 & case 2, take the max
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return max(maxSubArraySum, totalSum + minSubArraySum);
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}
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}

‎MaximumSumCircularSubarray/maximum_sum_circular_subarray.go

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package main
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func maxSubarraySumCircular(nums []int) int {
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var totalSum int = 0
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var maxSum int = -30012
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var curMax int = 0
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var minSum int = 3012
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var curMin = 0
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for _, x := range nums {
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curMax = max(x, curMax+x)
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maxSum = max(maxSum, curMax)
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curMin = min(x, curMin+x)
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minSum = min(minSum, curMin)
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totalSum += x
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}
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if maxSum > 0 {
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return max(maxSum, totalSum-minSum)
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} else {
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return maxSum
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}
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}
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func max(a int, b int) int {
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if a > b {
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return a
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} else {
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return b
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}
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}
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func min(a int, b int) int {
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if a < b {
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return a
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} else {
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return b
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}
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}
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/*
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int totalSum = 0, maxSum = -30012, curMax = 0, minSum = 30012, curMin = 0;
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for (int x in nums) {
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curMax = max(x, curMax + x); //update the current max sub-array sum
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maxSum = max(maxSum, curMax); //update the overall max sub-array sum
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curMin = min(x, curMin + x); //update the current min sub-array sum
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minSum = min(minSum, curMin); //update the overall min sub-array sum
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totalSum += x;
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}
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return maxSum > 0 ? max(maxSum, totalSum - minSum) : maxSum;
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*/

‎MaximumSumCircularSubarray/maximum_sum_circular_subarray.md

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# 🔥 100% FAST KADANE 🔥 || Simple Fast and Easy || with Explanation
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## Approach
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This problem is asking for the maximum possible sum of a non-empty sub-array of a given circular integer array.
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A circular array means that the end of the array connects to the beginning of the array, so the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].
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A sub-array may only include each element of the fixed buffer nums at most once.
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One approach to solving this problem is to first find the maximum sub-array sum using the standard Kadane's algorithm, and then check if there's a better solution by finding the minimum sub-array sum and subtracting it from the total sum of the array.
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This approach considers the case where the maximum sub-array sum is obtained from the sub-array that "wraps around" the circular array.
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### Complexity
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#### Time complexity:: O(n)
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Because we are iterating through the array once and performing constant time operations inside the loop
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#### Space complexity:: O(1)
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Because we are using a constant amount of additional space to store the curMax, maxSum, curMin, minSum and totalSum variables.
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## Code
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```dart
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class Solution {
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int maxSubarraySumCircular(List<int> nums) {
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int totalSum = 0, maxSum = -30012, curMax = 0, minSum = 30012, curMin = 0;
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for (int x in nums) {
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curMax = max(x, curMax + x); //update the current max sub-array sum
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maxSum = max(maxSum, curMax); //update the overall max sub-array sum
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curMin = min(x, curMin + x); //update the current min sub-array sum
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minSum = min(minSum, curMin); //update the overall min sub-array sum
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totalSum += x;
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}
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return maxSum > 0 ? max(maxSum, totalSum - minSum) : maxSum;
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}
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}
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```
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## Code - 2
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```dart
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class Solution {
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int kadane(List<int> nums) {
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int local = nums[0];
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int global = nums[0];
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for (int i = 1; i < nums.length; i++) {
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local = max(nums[i], local + nums[i]);
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global = max(global, local);
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}
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return global;
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}
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// case 1: max subarray sum in [0 .. n - 1]
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// i.e. kadane's algo
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// case 2. circular subarray in [0 .. | n - 1 .. | .. 2 * n - 1]
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// i.e. total sum - min subarray sum in [0 .. n - 1]
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int maxSubarraySumCircular(List<int> nums) {
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int n = nums.length;
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// use kadane's algo to find out max sub array sum (case 1)
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int maxSubArraySum = kadane(nums);
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// handle cases like [-3,-2,-3]
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if (maxSubArraySum < 0) return maxSubArraySum;
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// calculate the total sum
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int totalSum = 0;
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// in order to use the same kadane function, we flip the sign
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for (int i = 0; i < n; i++) {
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totalSum += nums[i];
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nums[i] = -nums[i];
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}
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// use kadane's algo to find out min sub array sum
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int minSubArraySum = kadane(nums);
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// compare case 1 & case 2, take the max
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return max(maxSubArraySum, totalSum + minSubArraySum);
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}
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}
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```

‎README.md

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- [**2421.** Number of Good Paths](NumberOfGoodPaths/number_of_good_paths.dart)
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- [**57.** Insert Interval](InsertInterval/insert_interval.dart)
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- [**926.** Flip String to Monotone Increasing](FlipStringToMonotoneIncreasing/flip_string_to_monotone_increasing.dart)
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- [**918.** Maximum Sum Circular Sub-Array](MaximumSumCircularSubarray/maximum_sum_circular_subarray.dart)
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## Reach me via
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