CPMagical CP Library

Problem solution · Java

Sum of K Subarrays With Length at Least M

Sum of K Subarrays With Length at Least M: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
37 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Sum of K Subarrays With Length at Least M, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 37 lines of Java from the credited upstream file 3473.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSum of K Subarrays With Length at Least M · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maxSum(int[] nums, int k, int m) {    final int n = nums.length;    int[] prefix = new int[n + 1];    Integer[][][] mem = new Integer[n][2][k + 1];    for (int i = 0; i < n; i++)      prefix[i + 1] = prefix[i] + nums[i];    return maxSum(nums, 0, 0, k, m, prefix, mem);  }   private static final int INF = 20_000_000;   private int maxSum(int[] nums, int i, int ongoing, int k, int m, int[] prefix,                     Integer[][][] mem) {    if (k < 0)      return -INF;    if (i == nums.length)      return k == 0 ? 0 : -INF;    if (mem[i][ongoing][k] != null)      return mem[i][ongoing][k];    if (ongoing == 1)      // 1. End the current subarray (transition to state 0, same index i)      // 2. Extend the current subarray by picking nums[i] and move to i + 1.      return mem[i][1][k] = Math.max(maxSum(nums, i, 0, k, m, prefix, mem),                                     maxSum(nums, i + 1, 1, k, m, prefix, mem) + nums[i]);    // ongoing == 0    // 1. Skip nums[i].    // 2. Pick nums[i..i + m - 1] (only if k > 0 and there're enough elements).    int res = maxSum(nums, i + 1, 0, k, m, prefix, mem);    if (i + m <= nums.length) // If we have enough elements for a new segment      res = Math.max(res,                     maxSum(nums, i + m, 1, k - 1, m, prefix, mem) + (prefix[i + m] - prefix[i]));     return mem[i][0][k] = res;  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗