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dpMaximumSubarray.js
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dpMaximumSubarray.js
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/**
* Dynamic Programming solution.
* Complexity: O(n)
*
* @param {Number[]} inputArray
* @return {Number[]}
*/
export default function dpMaximumSubarray(inputArray) {
// Check if all elements of inputArray are negative ones and return the highest
// one in this case.
let allNegative = true;
let highestElementValue = null;
for (let i = 0; i < inputArray.length; i += 1) {
if (inputArray[i] >= 0) {
allNegative = false;
}
if (highestElementValue === null || highestElementValue < inputArray[i]) {
highestElementValue = inputArray[i];
}
}
if (allNegative && highestElementValue !== null) {
return [highestElementValue];
}
// Let's assume that there is at list one positive integer exists in array.
// And thus the maximum sum will for sure be grater then 0. Thus we're able
// to always reset max sum to zero.
let maxSum = 0;
// This array will keep a combination that gave the highest sum.
let maxSubArray = [];
// Current sum and subarray that will memoize all previous computations.
let currentSum = 0;
let currentSubArray = [];
for (let i = 0; i < inputArray.length; i += 1) {
// Let's add current element value to the current sum.
currentSum += inputArray[i];
if (currentSum < 0) {
// If the sum went below zero then reset it and don't add current element to max subarray.
currentSum = 0;
// Reset current subarray.
currentSubArray = [];
} else {
// If current sum stays positive then add current element to current sub array.
currentSubArray.push(inputArray[i]);
if (currentSum > maxSum) {
// If current sum became greater then max registered sum then update
// max sum and max subarray.
maxSum = currentSum;
maxSubArray = currentSubArray.slice();
}
}
}
return maxSubArray;
}