Explore the Maximum Subarray problem with a brute-force triple loop, a quadratic improvement, and the optimal Kadane\u2019s Algorithm, all with intuition, fully-commented Python, dry runs, and complexity analysis.<\/p>\n\n\n\n
Here’s the [Problem Link<\/span><\/mark><\/a><\/strong>] to begin with.<\/p>\n\n\n Find the contiguous slice of an array whose elements add up to the largest possible sum.<\/em><\/p>\n\n\n\n We\u2019ll walk through three distinct solutions<\/strong>, each in its own self-contained section so you can master them one at a time:<\/p>\n\n\n\nSection<\/th> Time<\/th> Space<\/th> Core idea<\/th><\/tr><\/thead> 1. Brute Force (Triple Loop)<\/td> O(n\u00b3)<\/code><\/td>O(1)<\/code><\/td>Check every subarray, add fresh each time<\/td><\/tr> 2. Better Brute (Double Loop)<\/td> O(n\u00b2)<\/code><\/td>O(1)<\/code><\/td>Carry a running sum inside the inner loop<\/td><\/tr> 3. Optimal \u2013 Kadane<\/td> O(n)<\/code><\/td>O(1)<\/code><\/td>Drop negative baggage as you scan<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\nCommon Example for All Sections<\/h2>\n\n\n\n
nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]<\/code>
The best subarray is [4, -1, 2, 1]<\/code> with sum 6<\/strong>.<\/p>\n\n\n\n
\n\n\n\n1. Brute Force \u2013 Triple Loop<\/h2>\n\n\n\n
1.1 Intuition & Approach<\/h3>\n\n\n\n
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i<\/code>.<\/li>\n\n\n\nj \u2265 i<\/code>.<\/li>\n\n\n\ni<\/code> to j<\/code> in a third loop.<\/li>\n\n\n\n1.2 Python Code (with comments)<\/h3>\n\n\n\n