Given an input array that may contain both positive and negative integers, find the sum of continuous subarray of numbers which has the largest sum.
Example:
Input: nums = [-2, -5, 6, -2, -3, 1, 5, -6]
Output: 7
Explanation: sum([6, -2, -3, 1, 5]) = 7
Your algorithm should run in O(n) time and use O(1) extra space.
The problem requires finding the maximum sum of a continuous subarray within a given array of integers, which may include both positive and negative numbers.
nums
.Input: nums = [-2, -5, 6, -2, -3, 1, 5, -6]
Output: 7
Explanation: sum([6, -2, -3, 1, 5]) = 7
The core challenge is to identify the subarray with the maximum sum efficiently. This problem is significant in various applications such as financial analysis, where one might need to find the period with the highest profit.
Potential pitfalls include misunderstanding the requirement for a continuous subarray and not accounting for negative numbers correctly.
To solve this problem, we can use Kadane's Algorithm, which is an efficient way to find the maximum sum subarray in linear time.
A naive approach would involve checking all possible subarrays and calculating their sums, which would result in a time complexity of O(n^2). This is not optimal for large arrays.
Kadane's Algorithm improves upon the naive approach by maintaining a running sum of the current subarray and updating the maximum sum found so far. This approach ensures a time complexity of O(n) and space complexity of O(1).
Here is a step-by-step breakdown of Kadane's Algorithm:
maxCurrent
and maxGlobal
to the first element of the array.maxCurrent
to be the maximum of the current element and the sum of maxCurrent
and the current element.maxCurrent
is greater than maxGlobal
, update maxGlobal
.maxGlobal
will hold the maximum sum of the subarray./**
* Function to find the maximum sum of a continuous subarray
* @param {number[]} nums - The input array of integers
* @return {number} - The maximum sum of the subarray
*/
function maxSubArray(nums) {
// Initialize maxCurrent and maxGlobal to the first element
let maxCurrent = nums[0];
let maxGlobal = nums[0];
// Iterate through the array starting from the second element
for (let i = 1; i < nums.length; i++) {
// Update maxCurrent to be the maximum of the current element and the sum of maxCurrent and the current element
maxCurrent = Math.max(nums[i], maxCurrent + nums[i]);
// Update maxGlobal if maxCurrent is greater
if (maxCurrent > maxGlobal) {
maxGlobal = maxCurrent;
}
}
// Return the maximum sum found
return maxGlobal;
}
// Example usage:
const nums = [-2, -5, 6, -2, -3, 1, 5, -6];
console.log(maxSubArray(nums)); // Output: 7
The time complexity of Kadane's Algorithm is O(n) because it involves a single pass through the array. The space complexity is O(1) as it uses a constant amount of extra space.
Potential edge cases include:
Examples:
Input:[-1, -2, -3]
Output: -1 Input:[5]
Output: 5
To test the solution comprehensively, consider the following test cases:
Example test cases:
console.log(maxSubArray([-2, -5, 6, -2, -3, 1, 5, -6])); // Output: 7
console.log(maxSubArray([1, 2, 3, 4, 5])); // Output: 15
console.log(maxSubArray([-1, -2, -3, -4])); // Output: -1
console.log(maxSubArray([5])); // Output: 5
When approaching such problems, consider the following tips:
In this blog post, we discussed how to solve the Maximum Sum Subarray problem using Kadane's Algorithm. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving algorithmic thinking and coding skills.
For further reading and practice, consider the following resources:
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