Maximum amount subarm

Question

I am trying to find a subarray of contiguous ones in the array that has the largest amount. So for the array

{5, 15, -30, 10, -5, 40, 10}

the maximum amount possible using these numbers will be adjacent to 55, or (10 + (-5) + 40 + 10) = 55. In the program below, the maximum amount is 55, the problem I'm trying to figure out is how print the sequence that produces this 55. In other words, how can I print 10, -5, 40 and 10?

public static void main(String[] args) {
    int[] arr = {5, 15, -30, 10, -5, 40, 10};

    System.out.println(maxSubsequenceSum(arr));         
}

public static int maxSubsequenceSum(int[] X) {
    int max = X[0];
    int sum = X[0];

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

ArrayList sum i, ArrayList (5, 20, -10, 10, 5, 45, 55). ArrayList 0 , , , .

Answer 1

. max() sum max :

int sum_start = 0, sum_end = 0, start = 0, end = 0;
// In the for loop
if (X[i] > sum + X[i]) {
    sum = X[i];
    sum_start = i;
    sum_end = i;
} else {
    ++sum_end;
}
if (sum > max) {
    start = sum_start;
    end = sum_end;
    max = sum;
}

Answer 2

Math.Max ​​ if . :

    if X[i] > sum + X[i] then begin
        sum := X[i];
        start := i;
      end
      else
        sum := sum + X[i];
      if max < sum then begin
        max := sum;
        finish := i;
      end;

Answer 3

o (n) , reset , 0.

{5, 15, -30, 10, -5, 40, 10}

• 5 + 15 = 20
• 20 - 30 = -10 (reset )
• 10 -5 +40 +10 = 55
• . 55 -

edit: ... max,

• , u reset
• new max → ...

Answer 4

, :

package recursion;

import java.util.Arrays;

public class MaximumSubArray {

    private static SubArray maxSubArray(int[] values, int low, int high) {
        if (low == high) {
            // base condition
            return new SubArray(low, high, values[low]);
        } else {
            int mid = (int) (low + high) / 2;
            // Check left side
            SubArray leftSubArray = maxSubArray(values, low, mid);
            // Check right side
            SubArray rightSubArray = maxSubArray(values, mid + 1, high);
            // Check from middle
            SubArray crossSubArray = maxCrossSubArray(values, low, mid, high);
            // Compare left, right and middle arrays to find maximum sub-array
            if (leftSubArray.getSum() >= rightSubArray.getSum()
                    && leftSubArray.getSum() >= crossSubArray.getSum()) {
                return leftSubArray;
            } else if (rightSubArray.getSum() >= leftSubArray.getSum()
                    && rightSubArray.getSum() >= crossSubArray.getSum()) {
                return rightSubArray;
            } else {
                return crossSubArray;
            }
        }
    }

    private static SubArray maxCrossSubArray(int[] values, int low, int mid,
            int high) {
        int sum = 0;
        int maxLeft = low;
        int maxRight = high;

        int leftSum = Integer.MIN_VALUE;
        for (int i = mid; i >= low; i--) {
            sum = sum + values[i];
            if (sum > leftSum) {
                leftSum = sum;
                maxLeft = i;
            }
        }

        sum = 0;
        int rightSum = Integer.MIN_VALUE;
        for (int j = mid + 1; j <= high; j++) {
            sum = sum + values[j];
            if (sum > rightSum) {
                rightSum = sum;
                maxRight = j;
            }
        }
        SubArray max = new SubArray(maxLeft, maxRight, (leftSum + rightSum));
        return max;
    }

    static class SubArray {

        private int start;

        private int end;

        private int sum;

        public SubArray(int start, int end, int sum) {
            super();
            this.start = start;
            this.end = end;
            this.sum = sum;
        }

        public int getStart() { return start; }

        public void setStart(int start) { this.start = start; }

        public int getEnd() { return end; }

        public void setEnd(int end) { this.end = end; }

        public int getSum() { return sum; }

        public void setSum(int sum) { this.sum = sum; }

        @Override
        public String toString() {
            return "SubArray [start=" + start + ", end=" + end + ", sum=" + sum + "]";
        }
    }

    public static final void main(String[] args) {
        int[] values = { 5, 15, -30, 10, -5, 40, 10 };
        System.out.println("Maximum sub-array for array"
            + Arrays.toString(values) + ": " + maxSubArray(values, 0, 6));
    }

}

Maximum sub-array for array[5, 15, -30, 10, -5, 40, 10]: SubArray [start=3, end=6, sum=55]

https://github.com/gosaliajigar/Programs/blob/master/src/recursion/MaximumSubArray.java

Answer 5

, , , , i, . O (n) .

, , , .

Java, :

public int[] kadanesAlgorithm (int[] array) {
        int start_old = 0;
        int start = 0;
        int end = 0;
        int found_max = 0;

        int max = array[0];

        for(int i = 0; i<array.length; i++) {
            max = Math.max(array[i], max + array[i]);
            found_max = Math.max(found_max, max);
            if(max < 0)
                start = i+1;
            else if(max == found_max) {
                start_old=start;
                end = i;
                }
        }

        return Arrays.copyOfRange(array, start_old, end+1);
    }

Answer 6

.

public static int maxSubsequenceSum(int[] X) {
    int max = 0;
    boolean max_init = false;
    int max_from=0;
    int max_to=0; // this is not included
    for (int i = 0; i < X.length; i++) {
        for (int j = i + 1; j < X.length + 1; j++) {
            int total = 0;
            for (int k = i; k < j; k++) {
                total += X[k];
            }
            if (total > max || !max_init){
                max = total;
                max_init = true;
                max_from = i;
                max_to = j;
           }
        }
    }
    for (int i=max_from;i<max_to;i++)
       System.out.print(X[i]+",");
    System.out.println();
    return max;
}