Kadenes algorithim code simplified

Author: Arpit Sharma

Famous Algorithms/kadanes_algorithm.js

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+/* What is Kadane's Algorithm?
+Kadane's Algorithm is a way to find the maximum subarray sum in an array with a runtime of O(n).
+It is a dynamic programming algorithm that uses the fact that the maximum subarray sum ending at index i is either the value at index i or the maximum subarray sum ending at index i-1 plus the value at index i.
+
+Note: The subarray must be contiguous, all the values in the subarray must be next to each other in the original array.
+
+How does it work?
+In this algorithim we maintain two variables, one will hold the maximum sum of contagious subarray and the other variable will hold sum of next element + current sum from previous iteration.
+If at any point the current sum + next element sum is less then 0 then we will reset the current sum to 0 and current sum + next element sum will start again from the next element.
+If the current sum + next element sum is greater then 0 and it is also greater then the maximum sum of contagious subarray then the variable of maximum sum of contagious subarray will be updated with current sum + next element sum
+
+
+Lets take the example: {-2, -3, 4, -1, -2, 1, 5, -3}
+    In this example maxLargestSumTillNow holds the maximum sum of contagious subarray and newLargestSum hold the value of current sum + next element
+    On initalizing max so far will be the max -ve number
+    maxLargestSumTillNow = INT_MIN
+    newLargestSum = 0
+
+    for i=0,  a[0] =  -2
+    newLargestSum = newLargestSum + (-2)
+    Set newLargestSum = 0 because newLargestSum < 0
+    and set maxLargestSumTillNow = -2
+
+    for i=1,  a[1] =  -3
+    newLargestSum = newLargestSum + (-3)
+    Since newLargestSum = -3 and maxLargestSumTillNow = -2, maxLargestSumTillNow will remain -2
+    Set newLargestSum = 0 because newLargestSum < 0
+
+    for i=2,  a[2] =  4
+    newLargestSum = newLargestSum + (4)
+    newLargestSum = 4
+    maxLargestSumTillNow is updated to 4 because newLargestSum greater 
+    than maxLargestSumTillNow which was -2 till now
+
+    for i=3,  a[3] =  -1
+    newLargestSum = newLargestSum + (-1)
+    newLargestSum = 3
+
+    for i=4,  a[4] =  -2
+    newLargestSum = newLargestSum + (-2)
+    newLargestSum = 1
+
+    for i=5,  a[5] =  1
+    newLargestSum = newLargestSum + (1)
+    newLargestSum = 2
+
+    for i=6,  a[6] =  5
+    newLargestSum = newLargestSum + (5)
+    newLargestSum = 7
+    maxLargestSumTillNow is updated to 7 because newLargestSum is 
+    greater than maxLargestSumTillNow
+
+    for i=7,  a[7] =  -3
+    newLargestSum = newLargestSum + (-3)
+    newLargestSum = 4
+    
+    Time Complexity: O(n)
+    Space Complexity: O(1)
+*/
+
+function largestSumOfSubArray(arr) {
+    if (arr.lenth == 1) {
+        return arr[0];
+    }
+
+    // Variable for maintaining Maximum sum of the subarray
+    var maxint = Math.pow(2, 53);
+    var maxLargestSumTillNow = -maxint - 1;
+
+    // Variable to calclate the sum of subarray after each iteration
+    var newLargestSum = 0;
+
+    // Looping through the entire array
+    for (i = 0; i < arr.length - 1; i++) {
+        // Calculating the largest sum on each iteration
+        newLargestSum += arr[i];
+
+        // If the largest sum value is greater then the maximum largest subarray value we have maintained then we will assign new value to maintained maximum largest subarray
+        if (maxLargestSumTillNow < newLargestSum) {
+            maxLargestSumTillNow = newLargestSum;
+        }
+
+        // If the largest sum is negative then we will reset the value of largest sum to 0 and start the calculation again of largest sum from next element
+        if (newLargestSum < 0) {
+            newLargestSum = 0;
+        }
+    }
+
+    // After the completion of iteration we will return the max largest sub array value
+    return maxLargestSumTillNow;
+}
+
+// Driver code
+var arr = [-2, -3, 4, -1, -2, 1, 5, -3];
+console.log(largestSumOfSubArray(arr));
+
+// Input: arr = [-2, -3, 4, -1, -2, 1, 5, -3];
+//Output: 7