From 2ac4fd401ba199ecd4ae8d7fe97e6205fb19c0bd Mon Sep 17 00:00:00 2001
From: Leonid <l.khristyuk@smartagro.ru>
Date: Tue, 29 Apr 2025 14:19:17 +0300
Subject: [PATCH] =?UTF-8?q?Solved=20Maximum=20Subarray=20Sum=20(Kadane?=
 =?UTF-8?q?=E2=80=99s=20Algorithm)?=
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https://www.geeksforgeeks.org/largest-sum-contiguous-subarray/#naive-approach-by-iterating-over-all-subarrays-on-2-time-and-o1-space
---
 .../algos/codewars/MaximumSubarraySum.java    | 43 +++++++++++++++++++
 1 file changed, 43 insertions(+)
 create mode 100644 src/main/java/com/algos/codewars/MaximumSubarraySum.java

diff --git a/src/main/java/com/algos/codewars/MaximumSubarraySum.java b/src/main/java/com/algos/codewars/MaximumSubarraySum.java
new file mode 100644
index 0000000..ba33aba
--- /dev/null
+++ b/src/main/java/com/algos/codewars/MaximumSubarraySum.java
@@ -0,0 +1,43 @@
+package com.algos.codewars;
+//The maximum sum subarray problem consists in finding the maximum sum of a contiguous subsequence in an array
+// or list of integers:
+//
+//        Max.sequence(new int[]{-2, 1, -3, 4, -1, 2, 1, -5, 4});
+//// should be 6: {4, -1, 2, 1}
+//
+//        Easy case is when the list is made up of only positive numbers and the maximum sum is the sum
+//        of the whole array. If the list is made up of only negative numbers, return 0 instead.
+//
+//        Empty list is considered to have zero greatest sum. Note that the empty list or array is also
+//        a valid sublist/subarray.
+
+public class MaximumSubarraySum {
+
+    public static void main(String[] args) {
+        int [] testArray = new int[] {-2, 1, -3, 4, -1, 2, 1, -5, 4};
+//        int[] testArray = new int[]{2, 3, -8, 7, -1, 2, 3};
+        System.out.println(sequence(testArray));
+    }
+
+    public static int sequence(int[] arr) {
+
+        if (arr.length == 0) {
+            return 0;
+        }
+
+        int maxEnding = arr[0];
+        int result = arr[0];
+
+        for (int i = 1; i < arr.length; i++) {
+//            if (maxEnding < 0) {
+//                maxEnding = arr[i];
+//            } else {
+//                maxEnding + arr[i];
+//            }
+            maxEnding = Math.max(arr[i], maxEnding + arr[i]);
+            result = Math.max(result, maxEnding);
+        }
+
+        return Math.max(result, 0);
+    }
+}