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46 lines
1.4 KiB
JavaScript
46 lines
1.4 KiB
JavaScript
/**
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* Dynamic Programming solution.
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* Complexity: O(n)
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*
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* @param {Number[]} inputArray
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* @return {Number[]}
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*/
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export default function dpMaximumSubarray(inputArray) {
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// We iterate through the inputArray once, using a greedy approach to keep track of the maximum
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// sum we've seen so far and the current sum.
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//
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// The currentSum variable gets reset to 0 every time it drops below 0.
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//
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// The maxSum variable is set to -Infinity so that if all numbers are negative, the highest
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// negative number will constitute the maximum subarray.
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let maxSum = -Infinity;
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let currentSum = 0;
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// We need to keep track of the starting and ending indices that contributed to our maxSum
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// so that we can return the actual subarray. From the beginning let's assume that whole array
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// is contributing to maxSum.
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let maxStartIndex = 0;
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let maxEndIndex = inputArray.length - 1;
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let currentStartIndex = 0;
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inputArray.forEach((currentNumber, currentIndex) => {
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currentSum += currentNumber;
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// Update maxSum and the corresponding indices if we have found a new max.
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if (maxSum < currentSum) {
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maxSum = currentSum;
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maxStartIndex = currentStartIndex;
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maxEndIndex = currentIndex;
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}
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// Reset currentSum and currentStartIndex if currentSum drops below 0.
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if (currentSum < 0) {
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currentSum = 0;
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currentStartIndex = currentIndex + 1;
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}
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});
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return inputArray.slice(maxStartIndex, maxEndIndex + 1);
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}
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