MaximumSubarray.java

package problems.medium;

/**
 * Created by sherxon on 1/2/17.
 */
public class MaximumSubarray {

 /**
 * We use two max, maxSoFar -> max sub array found so far and maxEndingHere -> maximum sub array size
 * that ends in this index. This is in place solution and time Complexity is O(N).
 * */
 public int maxSubArray(int[] a) {

 if (a == null || a.length == 0) return 0;

 int maxSoFar = a[0];
 int maxEndingHere = a[0];

 for (int i = 1; i < a.length; i++) {
 maxEndingHere = Math.max(maxEndingHere + a[i], a[i]);
 maxSoFar = Math.max(maxSoFar, maxEndingHere);
 }
 return maxSoFar;
 }
}