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[Algorithm] Maximum Contiguous Subarray algorithm implementation using TypeScript / JavaScript

原创
05/13 14:22
阅读数 43226

Naive solution for this problem would be caluclate all the possible combinations:

const numbers = [1, -3, 2 - 5, 7, 6, -1, -4, 11, -23];

// O(n^3)
const findMaxSubAry = numbers => {
  let answer = Number.MIN_VALUE;
  /**
   * Calculate all the possible values and pick the max one
   * All possible values should be
   * length = 1, 2, ,3 ... n
   *  Pick differnet start point
   */

  // For different lenght
  for (let l = 0; l < numbers.length; l++) {
    // O(n)
    // For different start
    for (let s = 0; s < l; s++) {
      // O(n)
      if (s + l >= numbers.length) {
        break;
      }
      let sum = 0;
      for (let i = s; i < s + l; i++) {
        // O(n)
        sum += numbers[i];
      }

      answer = Math.max(answer, sum);
    }
  }

  return answer;
};

console.log(findMaxSubAry(numbers));  // 19

 

 

The maximum subarray problem is one of the nicest examples of dynamic programming application.

In this lesson we cover an example of how this problem might be presented and what your chain of thought should be to tackle this problem efficiently.

 /**
  * Maximum Contiguous subarray algorithm
  * 
  * Max(i) = Max(i-1) + v(i)
  * Max(i-1) < 0 ? v(i) : Max(i-1)
  * 
  * Combining
---------
maxInc(i) = maxInc(i - 1) > 0 ? maxInc(i - 1) + val(i) : val(i)
max(i) = maxInc(i) > max(i - 1) ? maxInc(i) : max(i - 1)
  */
function maxSumSubArray(arr) {
  /**
   *   inx  | val   |  max_inc    | max 
   *          0       0            0
   *    0     -2      0            0
   *    1     -3      0            0
   *    2     4       4            4     ---> start = 2
   *    3     -1      3            4
   *    4     -2      1            4
   *    5     1       2            4
   *    6     5       7            7     ---> end  = 6
   *    7     -3      4            7             
   */

  let val = 0, max_inc = 0, max = 0, start = 0, end = 0;

  for (let i = 1; i < arr.length; i++) {
    val = arr[i];
    max_inc = Math.max(max_inc + val, val);
    max = Math.max(max, max_inc);

    if (val === max_inc) {
      start = i;
    }

    if (max === max_inc) {
      end = i;
    }
  }

  if (end === 0) {
    end = start;
  }
  console.log(start, end);
  return arr.slice(start, end + 1);
}

console.log(maxSumSubArray([-2, -3, 4, -1, -2, 1, 5, -3])); //[4, -1, -2, 1, 5]
console.log(maxSumSubArray([-2,-3,-4,-1,-2])); // [-2]

 

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