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* Add Tug of War solution using backtracking * Updated Documentation in README.md * Added Tug of war problem link * Updated Documentation in README.md * Updated Documentation in README.md * Refactor tugOfWar: remove unused vars, optimize initialization, and remove redundant checks * Added Function Export Statment * Updated Documentation in README.md * Resolved Code Style --Prettier * Rename "backtrack" to "recurse" * Fix test case: The difference needs to be exactly 1. * Code Modification: subsets should have sizes as close to n/2 as possible * Updated test-case of TugOfWar * Changed TugOfWar problem to Partition * Modified partition problem * Updated Documentation in README.md * fixed code style --------- Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com> Co-authored-by: Lars Müller <34514239+appgurueu@users.noreply.github.com>
40 lines
1.2 KiB
JavaScript
40 lines
1.2 KiB
JavaScript
/**
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* @function canPartition
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* @description Check whether it is possible to partition the given array into two equal sum subsets using recursion.
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* @param {number[]} nums - The input array of numbers.
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* @param {number} index - The current index in the array being considered.
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* @param {number} target - The target sum for each subset.
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* @return {boolean}.
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* @see [Partition Problem](https://en.wikipedia.org/wiki/Partition_problem)
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*/
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const canPartition = (nums, index = 0, target = 0) => {
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if (!Array.isArray(nums)) {
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throw new TypeError('Invalid Input')
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}
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const sum = nums.reduce((acc, num) => acc + num, 0)
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if (sum % 2 !== 0) {
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return false
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}
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if (target === sum / 2) {
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return true
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}
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if (index >= nums.length || target > sum / 2) {
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return false
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}
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// Include the current number in the first subset and check if a solution is possible.
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const withCurrent = canPartition(nums, index + 1, target + nums[index])
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// Exclude the current number from the first subset and check if a solution is possible.
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const withoutCurrent = canPartition(nums, index + 1, target)
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return withCurrent || withoutCurrent
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}
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export { canPartition }
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