[리트코드] 53. Maximum Subarray (DP, Kadane's algorithm)

문제 정보

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

A subarray is a contiguous part of an array.

 

문제 풀이

단순하게 모든 부분합을 반복하며 최대값을 구해 저장할 수 도 있겠지만 이는 O(n^2) 시간복잡도를 가진다.

DP나 카데인 알고리즘을 사용하여 풀이하면 O(n)으로 해결할 수 있다.

 

Kadane's algorithm 이란?

The simple idea of Kadane’s algorithm is to look for all positive contiguous segments of the array (max_ending_here is used for this). And keep track of maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive-sum compare it with max_so_far and update max_so_far if it is greater than max_so_far 

 

Kadane's algorithm은 DP알고리즘의 일종으로 부분합의 최대값(Local maximum)에서 이후 엔트리가 양인 경우만 새로운 local maximum으로 지정해가면서 값을 갱신해나가면 된다.

 

Kadane's algorithm 

Initialize:
    max_so_far = INT_MIN
    max_ending_here = 0

Loop for each element of the array
  (a) max_ending_here = max_ending_here + a[i]
  (b) if(max_so_far < max_ending_here)
            max_so_far = max_ending_here
  (c) if(max_ending_here < 0)
            max_ending_here = 0
return max_so_far

The simple idea of Kadane’s algorithm is to look for all positive contiguous segments of the array (max_ending_here is used for this). And keep track of maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive-sum compare it with max_so_far and update max_so_far if it is greater than max_so_far

    Lets take the example:
    {-2, -3, 4, -1, -2, 1, 5, -3}

    max_so_far = max_ending_here = 0

    for i=0,  a[0] =  -2
    max_ending_here = max_ending_here + (-2)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=1,  a[1] =  -3
    max_ending_here = max_ending_here + (-3)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=2,  a[2] =  4
    max_ending_here = max_ending_here + (4)
    max_ending_here = 4
    max_so_far is updated to 4 because max_ending_here greater 
    than max_so_far which was 0 till now

    for i=3,  a[3] =  -1
    max_ending_here = max_ending_here + (-1)
    max_ending_here = 3

    for i=4,  a[4] =  -2
    max_ending_here = max_ending_here + (-2)
    max_ending_here = 1

    for i=5,  a[5] =  1
    max_ending_here = max_ending_here + (1)
    max_ending_here = 2

    for i=6,  a[6] =  5
    max_ending_here = max_ending_here + (5)
    max_ending_here = 7
    max_so_far is updated to 7 because max_ending_here is 
    greater than max_so_far

    for i=7,  a[7] =  -3
    max_ending_here = max_ending_here + (-3)
    max_ending_here = 4

Writeup (using DP)

public class Solution
{
    public int MaxSubArray(int[] nums)
    {
        int[] local_maximum = new int[nums.Length];
        int global_maximum;

        global_maximum = nums[0];
        local_maximum[0] = nums[0];
        
        for (int i = 0; i < nums.Length; i++)
        {
            if (i != 0)
            {
                local_maximum[i] = local_maximum[i - 1] + nums[i];
            }

            if (local_maximum[i] > global_maximum)
            {
                global_maximum = local_maximum[i];
            }

            if (local_maximum[i] < 0)
            {
                local_maximum[i] = 0;
            }
        }

        return global_maximum;
    }

}

 

Writeup (using Kadane's algorithm)

static int maxSubArraySum(int[] a, int size)
{
    int max_so_far = a[0], max_ending_here = 0;
 
    for (int i = 0; i < size; i++) {
        max_ending_here = max_ending_here + a[i];
        if (max_ending_here < 0)
            max_ending_here = 0;
 
        /* Do not compare for all
        elements. Compare only
        when max_ending_here > 0 */
        else if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    return max_so_far;
}

 

 

 

출처 : https://www.geeksforgeeks.org/largest-sum-contiguous-subarray/

Largest Sum Contiguous Subarray - GeeksforGeeks