Swift Knapsack problem using bottom-down (memoization) approach












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$begingroup$


I'm solving the "unbounded" variant of the knapsack problem, meaning the repetition of items is allowed. As in the Hackerrank version of the knapsack problem, I am returning the sum nearest to, not exceeding the target.



Some sample calls, with the solutions afterwards ar



unboundedKnapsack(k: 10, arr: [2,3,4]) // 10
unboundedKnapsack(k: 12, arr: [1,6,9]) //12
unboundedKnapsack(k: 9, arr: [3,4,4,4,8]) // 9
unboundedKnapsack(k: 3, arr: [2]) // 2
unboundedKnapsack(k: 13, arr: [3,7,9,11]) //13
unboundedKnapsack(k: 11, arr: [3,7,9]) // 10


I developed a recursive solution, and as below I've added memoization:



func unboundedKnapsack(k: Int, arr: [Int]) -> Int {
var cache: [[Int]] = Array(repeating: Array(repeating: 0, count: arr.count), count: k)
return k - knap(k, arr, 0, 0, &cache)
}

func knap(_ target: Int, _ arr: [Int], _ ptr: Int, _ current : Int, _ cache: inout [[Int]]) -> Int {
// greedy - we either take the current item or we don't
if (current > target) {return Int.max}
if (ptr > arr.count - 1) {return target - current}
if (current == target) {return target - current}
if (cache[current][ptr] != 0) {return cache[current][ptr]}
let result = min(
// take the current and move pointer
knap(target, arr, ptr + 1, current + arr[ptr], &cache)
,
// take the current and leave the pointer so we can take more
knap(target, arr, ptr, current + arr[ptr], &cache),
// do not take the current and move pointer
knap(target, arr, ptr + 1, current, &cache)
)
cache[current][ptr] = result
return result
}


I haven't been able to test all inputs for my solution, so I wanted comments on how this particular approach could be improved (using my memoization as shown rather than changing the approach).









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    0












    $begingroup$


    I'm solving the "unbounded" variant of the knapsack problem, meaning the repetition of items is allowed. As in the Hackerrank version of the knapsack problem, I am returning the sum nearest to, not exceeding the target.



    Some sample calls, with the solutions afterwards ar



    unboundedKnapsack(k: 10, arr: [2,3,4]) // 10
    unboundedKnapsack(k: 12, arr: [1,6,9]) //12
    unboundedKnapsack(k: 9, arr: [3,4,4,4,8]) // 9
    unboundedKnapsack(k: 3, arr: [2]) // 2
    unboundedKnapsack(k: 13, arr: [3,7,9,11]) //13
    unboundedKnapsack(k: 11, arr: [3,7,9]) // 10


    I developed a recursive solution, and as below I've added memoization:



    func unboundedKnapsack(k: Int, arr: [Int]) -> Int {
    var cache: [[Int]] = Array(repeating: Array(repeating: 0, count: arr.count), count: k)
    return k - knap(k, arr, 0, 0, &cache)
    }

    func knap(_ target: Int, _ arr: [Int], _ ptr: Int, _ current : Int, _ cache: inout [[Int]]) -> Int {
    // greedy - we either take the current item or we don't
    if (current > target) {return Int.max}
    if (ptr > arr.count - 1) {return target - current}
    if (current == target) {return target - current}
    if (cache[current][ptr] != 0) {return cache[current][ptr]}
    let result = min(
    // take the current and move pointer
    knap(target, arr, ptr + 1, current + arr[ptr], &cache)
    ,
    // take the current and leave the pointer so we can take more
    knap(target, arr, ptr, current + arr[ptr], &cache),
    // do not take the current and move pointer
    knap(target, arr, ptr + 1, current, &cache)
    )
    cache[current][ptr] = result
    return result
    }


    I haven't been able to test all inputs for my solution, so I wanted comments on how this particular approach could be improved (using my memoization as shown rather than changing the approach).









    share









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm solving the "unbounded" variant of the knapsack problem, meaning the repetition of items is allowed. As in the Hackerrank version of the knapsack problem, I am returning the sum nearest to, not exceeding the target.



      Some sample calls, with the solutions afterwards ar



      unboundedKnapsack(k: 10, arr: [2,3,4]) // 10
      unboundedKnapsack(k: 12, arr: [1,6,9]) //12
      unboundedKnapsack(k: 9, arr: [3,4,4,4,8]) // 9
      unboundedKnapsack(k: 3, arr: [2]) // 2
      unboundedKnapsack(k: 13, arr: [3,7,9,11]) //13
      unboundedKnapsack(k: 11, arr: [3,7,9]) // 10


      I developed a recursive solution, and as below I've added memoization:



      func unboundedKnapsack(k: Int, arr: [Int]) -> Int {
      var cache: [[Int]] = Array(repeating: Array(repeating: 0, count: arr.count), count: k)
      return k - knap(k, arr, 0, 0, &cache)
      }

      func knap(_ target: Int, _ arr: [Int], _ ptr: Int, _ current : Int, _ cache: inout [[Int]]) -> Int {
      // greedy - we either take the current item or we don't
      if (current > target) {return Int.max}
      if (ptr > arr.count - 1) {return target - current}
      if (current == target) {return target - current}
      if (cache[current][ptr] != 0) {return cache[current][ptr]}
      let result = min(
      // take the current and move pointer
      knap(target, arr, ptr + 1, current + arr[ptr], &cache)
      ,
      // take the current and leave the pointer so we can take more
      knap(target, arr, ptr, current + arr[ptr], &cache),
      // do not take the current and move pointer
      knap(target, arr, ptr + 1, current, &cache)
      )
      cache[current][ptr] = result
      return result
      }


      I haven't been able to test all inputs for my solution, so I wanted comments on how this particular approach could be improved (using my memoization as shown rather than changing the approach).









      share









      $endgroup$




      I'm solving the "unbounded" variant of the knapsack problem, meaning the repetition of items is allowed. As in the Hackerrank version of the knapsack problem, I am returning the sum nearest to, not exceeding the target.



      Some sample calls, with the solutions afterwards ar



      unboundedKnapsack(k: 10, arr: [2,3,4]) // 10
      unboundedKnapsack(k: 12, arr: [1,6,9]) //12
      unboundedKnapsack(k: 9, arr: [3,4,4,4,8]) // 9
      unboundedKnapsack(k: 3, arr: [2]) // 2
      unboundedKnapsack(k: 13, arr: [3,7,9,11]) //13
      unboundedKnapsack(k: 11, arr: [3,7,9]) // 10


      I developed a recursive solution, and as below I've added memoization:



      func unboundedKnapsack(k: Int, arr: [Int]) -> Int {
      var cache: [[Int]] = Array(repeating: Array(repeating: 0, count: arr.count), count: k)
      return k - knap(k, arr, 0, 0, &cache)
      }

      func knap(_ target: Int, _ arr: [Int], _ ptr: Int, _ current : Int, _ cache: inout [[Int]]) -> Int {
      // greedy - we either take the current item or we don't
      if (current > target) {return Int.max}
      if (ptr > arr.count - 1) {return target - current}
      if (current == target) {return target - current}
      if (cache[current][ptr] != 0) {return cache[current][ptr]}
      let result = min(
      // take the current and move pointer
      knap(target, arr, ptr + 1, current + arr[ptr], &cache)
      ,
      // take the current and leave the pointer so we can take more
      knap(target, arr, ptr, current + arr[ptr], &cache),
      // do not take the current and move pointer
      knap(target, arr, ptr + 1, current, &cache)
      )
      cache[current][ptr] = result
      return result
      }


      I haven't been able to test all inputs for my solution, so I wanted comments on how this particular approach could be improved (using my memoization as shown rather than changing the approach).







      swift





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      asked 4 mins ago









      stevenpcurtisstevenpcurtis

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