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Java Array Exercises: Find contiguous subarray within a given array of integers which has the largest sum

Java Array: Exercise-66 with Solution

Write a Java program to find contiguous subarray within a given array of integers which has the largest sum.

In computer science, the maximum sum subarray problem is the task of finding a contiguous subarray with the largest sum, within a given one-dimensional array A[1...n] of numbers. Formally, the task is to find indices and with, such that the sum is as large as possible.

Example:
Input :
int[] A = {1, 2, -3, -4, 0, 6, 7, 8, 9}
Output:
The largest sum of contiguous sub-array: 30

Sample Solution:

Java Code:

import java.util.Arrays;
class solution	
{
	public static int largest_sum(int[] A)
	{
		int max_ele_val = 0;
		int max_end = 0;
		for (int i: A)
		{
			max_end = max_end + i;
			max_end = Integer.max(max_end, 0);

			max_ele_val = Integer.max(max_ele_val, max_end);
		}
		return max_ele_val;
	}
	public static void main(String[] args)
	{
		int[] A = {1, 2, -3, -4, 0, 6, 7, 8, 9};
		System.out.println("\nOriginal array: "+Arrays.toString(A));
		System.out.println("The largest sum of contiguous sub-array: " + largest_sum(A));
	}
}

Sample Output:

Original array: [1, 2, -3, -4, 0, 6, 7, 8, 9]
The largest sum of contiguous sub-array: 30

Flowchart:

Flowchart: Find contiguous subarray within a given array of integers which has the largest sum.

Java Code Editor:

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Previous: Write a Java program to find maximum difference between two elements in a given array of integers such that smaller element appears before larger element.
Next: Write a Java program to find subarray which has the largest sum in a given circular array of integers.

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Java: Tips of the Day

countOccurrences

Counts the occurrences of a value in an array.

Use Arrays.stream().filter().count() to count total number of values that equals the specified value.

public static long countOccurrences(int[] numbers, int value) {
    return Arrays.stream(numbers)
            .filter(number -> number == value)
            .count();
}

Ref: https://bit.ly/3kCAgLb