Cfxtrjtrk

You could iterate the array and take either the next element or if no element is taken before omit this element.

function getSubset(array, sum) {

    function iter(temp, delta, index) {

        if (!delta) result.push(temp);

        if (index >= array.length) return;

        iter(temp.concat(array[index]), delta - array[index], index + 1);

        if (!temp.length) iter(temp, delta, index + 1);

    }



    var result = ;

    iter(, sum, 0);

    return result;

}



console.log(getSubset([2, 4, 45, 6, 0, 19], 51));                   // [2, 4, 45], [45, 6], [45, 6, 0]

console.log(getSubset([1, 11, 100, 1, 0, 200, 3, 2, 1, 280], 280)); // [280]

console.log(getSubset([1, 3, 6, 11, 1, 5, 4], 4));                  // [1, 3], [4]

It will give all the available case. And I use the test case of @Nina Scholz

const sum = arr => arr.reduce((a,b) => a + b)



function cal(arr, x) {

  const rs = 

  for (let i = 0; i< arr.length; i++) {

    const tmp = 

    for (let j=i; j<arr.length; j++ ) {

      tmp.push(arr[j])

      if(sum(tmp) === x) rs.push([...tmp])

    }

  }

  return rs

}





console.log(cal([1, 11, 100, 1, 0, 200, 3, 2, 1, 280], 280)) // -> [280]

console.log(cal([2, 4, 45, 6, 0, 19], 51)); // -> [2, 4, 45] [45, 6] [45, 6, 0]

console.log(cal([1, 3, 6, 11, 1, 5, 4], 4)); // -> [1,3] [4]

This will try every possible permutation of the array (will stop further permutations once limit is reached)

function test(arr, num) {

  // sorting will improve time as larger values will be eliminated first

  arr = arr.sort(function(a, b) {

    return b - a;

  });

  var allLists = ;

  var start = Date.now();

  helper(0, 0, );

  console.log("Ms elapesed: " + (Date.now() - start));

  return allLists || "Not found";



  function helper(start, total, list) {

    var result = ;

    // Using for loop is faster because you can start from desired index without using filter, slice, splice ...

    for (var index = start; index < arr.length; index++) {

      var item = arr[index];

      // If the total is too large the path can be skipped alltogether

      if (total + item <= num) {

        // Check lists if number was not included

        var test = helper(index + 1, total, list.concat(result)); // remove for efficiency

        total += item;

        result.push(item);

        //if (total === num) index = arr.length; add for efficiency

      }

    }

    if (total === num) allLists.push(list.concat(result));

  }

}







console.log(test([2, 4, 45, 6, 0, 19], 51)); // [2,4,45] [2,4,45,0] [6,45] [6,45,0]

console.log(test([1, 11, 100, 1, 0, 200, 3, 2, 1, 280], 280)); // [280] [280,0]

If you want to make it more efficient and just return one of the resulted array just comment out the recursive call. You can also un-comment the line that exits the loop once the limit has been reached (will skip 0s).

If the question is about finding all subsets (rather than subarrays) with the given cross sum it is also known as the perfect sum problem.
https://www.geeksforgeeks.org/perfect-sum-problem-print-subsets-given-sum/

// A recursive function to print all subsets with the 

// help of dp. Vector p stores current subset. 

function printSubsetsRec(arr, i, sum, p) 

{ 

    // If we reached end and sum is non-zero. We print 

    // p only if arr[0] is equal to sun OR dp[0][sum] 

    // is true. 

    if (i == 0 && sum != 0 && dp[0][sum]) 

    { 

        p.push(arr[i]); 

        console.log(p); 

        return; 

    } 



    // If sum becomes 0 

    if (i == 0 && sum == 0) 

    { 

        console.log(p); 

        return; 

    } 



    // If given sum can be achieved after ignoring 

    // current element. 

    if (dp[i-1][sum]) 

    { 

        // Create a new vector to store path 

        var b = p.slice(0); 

        printSubsetsRec(arr, i-1, sum, b); 

    } 



    // If given sum can be achieved after considering 

    // current element. 

    if (sum >= arr[i] && dp[i-1][sum-arr[i]]) 

    { 

        p.push(arr[i]); 

        printSubsetsRec(arr, i-1, sum-arr[i], p); 

    } 

} 



// Prints all subsets of arr[0..n-1] with sum 0. 

function printAllSubsets(arr, sum) 

{ 

    var n = arr.length

    if (n == 0 || sum < 0) 

       return; 



    // Sum 0 can always be achieved with 0 elements 

    dp = ; 

    for (var i=0; i<n; ++i) 

    { 

        dp[i] = 

        dp[i][0] = true; 

    } 



    // Sum arr[0] can be achieved with single element 

    if (arr[0] <= sum) 

       dp[0][arr[0]] = true; 



    // Fill rest of the entries in dp 

    for (var i = 1; i < n; ++i) 

        for (var j = 0; j < sum + 1; ++j) 

            dp[i][j] = (arr[i] <= j) ? dp[i-1][j] || 

                                       dp[i-1][j-arr[i]] 

                                     : dp[i - 1][j]; 

    if (dp[n-1][sum] == false) 

    { 

        console.log("There are no subsets with sum %dn", sum); 

        return; 

    } 



    // Now recursively traverse dp to find all 

    // paths from dp[n-1][sum] 

    var p = ; 

    printSubsetsRec(arr, n-1, sum, p); 

} 



printAllSubsets([1,2,3,4,5], 10); 

Solution

'use strict';



function print(arr, i, j) {

   let k = 0;

   for (k = i; k <= j; k += 1) {

     console.log(arr[k]);

   }

}



function findSubArrays(arr, sum) {

  let n = arr.length;

  let i;

  let j;

  let sum_so_far;



  for (i = 0; i<n; i+= 1) {

    sum_so_far = 0;

    for (j = i; j < n; j++) {

      sum_so_far += arr[j];



      if (sum_so_far === sum) {

         print(arr, i, j);

      }

    }



  }

}

I would first loop depending on the size of expected arrays.

After that loop for looking for first part of the array which should be filled with positions that will match the desired number.

For example for x= 4 having arr=[5,4,32,8,2,1,2,2,3,4,4]
It would first take the 4's. Output will start on [ [4], [4], [4], ..... ] for positions 1,9,10 (respectively)

Then go for the arrays resulting sum of 2 elements [ ... [2,2], [2,2],[2,2], [1,3] ...] ( positions 4+6, position 4+7 position6+7 and position 5+8)
You would probably want to use another function to sum and check at this point.

Now will do the same for sum of 3 elements (if any) and so on, having max loop set at number of original array (the resulting number could be the sum of all the elements in the array).

The resulting example would be [ [4], [4], [4], [2,2], [2,2],[2,2], [1,3]]

function combinations(array) {

    return new Array(1 << array.length).fill().map(

        (e1,i) => array.filter((e2, j) => i & 1 << j));

}

function add(acc,a) {

  return acc + a 

}

combinations([2, 4, 45, 6, 0, 19]).filter( subarray => subarray.reduce(add, 0)  == 51 )

output

[[2,4,45],[45,6],[2,4,45,0],[45,6,0]]

combinations([1, 11, 100, 1, 0, 200, 3, 2, 1, 280]).filter( subarray => subarray.reduce(add, 0)  == 280 )

output

[[280],[0,280]]