Ggtkuk

This might not be exactly what's needed - might require tweaking as the logic may be flawed here.

I have commented the code for clarification.

var arr = [1, 3, 6, 11, 1, 5,4];	//  Define array



var target = 31;	//  Define target



//  filter the numbers higher than target and sort rest ascending

var withinRange = arr.filter(x => x <= target).sort((a, b) => a - b);

                      

if(arr.reduce((a,b) => a + b) < target)	//  Check if we have enough numbers to make up that number

  throw "The max you can get out of your selection is: " + arr.reduce((a,b) => a + b);

                      

//  grab the highest number as a starting point and remove it from our array of numbers

var numbers = [withinRange.pop()];



var toFind = target - getSum();	//  get remainder to find



for(var i = withinRange.length - 1; i > -1; i--)	//  iterate from the top

{



  if(toFind == withinRange[i]){	//  check if number is exactly what we need

    numbers.push(withinRange[i]);

    break;

  }else if(withinRange[i] <= toFind){	//  if number is smaller than what we look for

    numbers.push(withinRange[i]);

    toFind -= withinRange[i];

  }



}



function getSum(){	//  sum up our found numbers

  if(numbers.length == 0) return 0;

  return numbers.reduce((a,b) => a + b);

}



console.log([numbers, [target]]);	//  print numbers as desired output

console.log(target, getSum())	//  print the target and our numbers

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]]