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Maximum Subarray

Medium Day 1 • Striver Blind 75

Given an integer array nums, find the subarray with the largest sum, and return its sum.

Example 1:

  • Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
  • Output: 6
  • Explanation: The subarray [4,-1,2,1] has the largest sum 6.

Example 2:

  • Input: nums = [1]
  • Output: 1

Example 3:

  • Input: nums = [5,4,-1,7,8]
  • Output: 23

Constraints:

  • 1 ≤ nums.length ≤ 10⁵
  • -10⁴ ≤ nums[i] ≤ 10⁴

Maximum Subarray (Kadane’s Algorithm) is a classic DP problem that teaches optimal substructure.

Pattern: Kadane’s Algorithm

Track current subarray sum and maximum sum seen so far. Reset to 0 if current sum goes negative.


📊 Step-by-Step Execution (Mermaid Diagram)

Section titled “📊 Step-by-Step Execution (Mermaid Diagram)”
graph TD
Problem["Problem of Size N"] --> Sub["Break into Subproblems DP[i]"]
Sub --> Base["Base Cases: DP[0], DP[1]"]
Base --> Trans["State Transition: DP[i] = f(DP[i-1], DP[i-2], ...)"]
Trans --> Table["Fill DP Table / Variables"]
Table --> Result["Return DP[N]"]

function maxSubArray(nums) {
let max = -Infinity;
for (let i = 0; i < nums.length; i++) {
let sum = 0;
for (let j = i; j < nums.length; j++) {
sum += nums[j];
max = Math.max(max, sum);
}
}
return max;
}
  • Time Complexity: O(n²)
  • Space Complexity: O(1)
  • Explanation: Check every possible subarray.

function maxSubArray(nums) {
let maxSum = nums[0];
let currentSum = nums[0];
for (let i = 1; i < nums.length; i++) {
currentSum = Math.max(nums[i], currentSum + nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
  • Time Complexity: O(n)
  • Space Complexity: O(1)
  • Explanation: Kadane’s Algorithm.

  1. Initialize State: Setup necessary pointers, dynamic programming arrays, or hash maps.
  2. Iterate & Evaluate: Process the input according to the boundary conditions.
  3. Update & Return: Compute the optimal answer and return early or at termination.
  1. Start with brute force O(n²)
  2. Extending a negative prefix never helps
  3. Compare current element vs current element + previous sum
  4. Track global maximum

  1. What is the maximum sum ending at each position?
  2. If current sum goes negative, start fresh.
  3. Track two values: current sum and max sum.

👉 Solve this problem interactively in the DSA Lab