Maximum Subarray
Created: February 20, 2019 by [lek-tin]
Last updated: October 11, 2019
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
Follow up:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
Solution (Dynamic Programming)
class Solution {
public int maxSubArray(int[] nums) {
if (nums.length == 0) {
return 0;
}
int[] mem = new int[nums.length];
mem[0] = nums[0];
int max = nums[0];
for (int i = 1; i < nums.length; i++) {
mem[i] = mem[i-1] > 0 ? mem[i-1] + nums[i] : nums[i];
max = Math.max(mem[i], max);
}
return max;
}
}
Solution (prefix sum)
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
prefixSum, minSum, maxSum = 0, 0, -math.inf
for i, num in enumerate(nums):
# prefixSum[n+1] = prefixSum[n] + nums[n+1]
prefixSum += num
# maxSum = prefixSum[n+1] - min{prefixSum[0-->n]}
maxSum = max(prefixSum-minSum, maxSum)
# Update minSum now we have min{prefixSum[0-->n+1]}
# we only care when minSum is negative because so want to drop any negative preSum
minSum = min(prefixSum, minSum)
return maxSum
Need to return the max subarray
import math
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
preSums, minSum, maxSum = 0, 0, -math.inf
start, left, right = 0, 0, 0
for i, num in enumerate(nums):
preSums += num
if maxSum < preSums:
maxSum = preSums
left = start
right = i
if preSums < 0:
preSums = 0
start = i+1
return nums[left:right+1]
Solution (Kadane’s Algorithm)
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
n = len(nums)
if n == 0:
return 0
localMaxSum = globalMaxSum = nums[0]
for i in range(1, n):
# restart if previous local max sum is negative
localMaxSum = max(localMaxSum+nums[i], nums[i])
globalMaxSum = max(globalMaxSum, localMaxSum)
return globalMaxSum