找到等于所需总和的两个数字

Question

I had a technical interview and one of the questions was to find the two numbers in the array that would equal the desired sum. For the life of me, I can not find the solution.

 var input = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5]; var dSum = 28; // return would be [12,16]

Answer 1

Use a Set() , for solving this problem in O(n) . The approach is simple :

• Take an empty set. And for each element e in input check:

  (a) If the set contains sum - e . if yes then print the pair (e, sum -e) .

  (b) Insert e into set.

  Try the following:

 let input = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5]; let dSum = 28; let set = new Set(); for(item of input) { let num = dSum - item; if(set.has(num)) { console.log(num + " " + item); break; } set.add(item); } 

Answer 2

Please have a look at this simple code because I'm pretty sure this is what you need:

 function detectPair(sum, array) { for (i=0; i<array.length; i++) { for (j=0; j<array.length; j++) { if (i == j) continue; else if (array[i] + array[j] === sum) return [array[i], array[j]]; } }; return null; } let sum = 28; let array = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5] console.log(detectPair(sum, array)); //return would be [12,16] 

---

Edit @uom-pgregorio : ' There wasn't any constraints that the same number can't be used. So a 14 + 14 in this case should still be accepted There wasn't any constraints that the same number can't be used. So a 14 + 14 in this case should still be accepted '

I'm pretty sure you've mixed something up : The line if (i == j) continue; is not preventing the situation that 14+14 is correct, it is preventing that if the array just includes a number (like 14) it uses the same number a several times.

==> `i == j` is checking the indexes not the values

Maybe just try this setup :

let sum = 28;
let array = [3, 5, 7, 14, 14] // includes 14 two times

Answer 3

 function findSumPair(input, target) { for(let a = 0; a < input.length; a++) { for(let b = a; b < input.length; b++) { if(input[a]+input[b] === target) return [input[a], input[b]]; } } } console.log(findSumPair([3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5], 28)); 

The inner loop can be started at the current index of the outer loop (rather than 0) because the previous combinations have already been checked.

Answer 4

 var input = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5]; var dSum = 28; // return would be [12,16] let el1 = Math.max(...input) let el2 = Math.max(...input.filter(i => i !== el1)) console.log([el1, el2]) 

Or

 let input = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5]; let dSum = 28; // return would be [12,16] let result for (let i = 0; i < input.length; i++) for (let k = i; k < input.length; k++) if (i !== k && input[i] + input[k] === dSum) result = [input[i], input[k]] console.log(result) 

Answer 5

Beside the double nested solutions and solutions with more than one loop, you could use a single loop approach with a hash table for the still missing number and a short circuit if this number is found in the array.

This approach is a fast one, because it uses a for loop with an object as storage for a fast access, and a variable value for array[i] .

Big O is in worst case O(n).

 function getPair(array, sum) { var i, l, hash = Object.create(null), value; for (i = 0, l = array.length; i < l; i++) { value = array[i]; if (hash[value]) { return [sum - value, value]; } hash[sum - value] = true; } } console.log(getPair([3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5], 28)); 

Answer 6

I'd say this works, but the algorithm is really bad, O(n^2) I think.

for (i of input) {
    for (j of input) {
        if (i + j == dSum)
            console.log([i, j]);
    }
}

Answer 7

 function findPair(input, sum) { for(let i = 0; i < input.length; i++) { for(let j = i; j < input.length; j++) { if(input[i] + input[j] == sum) return [input[i], input[j]]; } } return []; } console.log(findPair([3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5], 28)); 

Answer 8

Try the following:

   for (var i = 0; i < input.length; i++)
    for (var j = 0; J < input.length; j++)
        if (input[i]+input[j] == dSum)
            console.log([input[i], input[j]])

Edit: I thought there was a constraint of i<>j. I removed that constraint

Answer 9

 var input = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5]; var dSum = 28; twoSum(input, dSum); function twoSum (arr, target) { let results = []; for (let i=0; i<arr.length; i++) { for (let j=i+1; j<arr.length; j++) { if (arr[j] === target - arr[i]) { results.push([arr[i], arr[j]]) } } } return results; } 

Answer 10

@amrender-singh Has the best answer here.

To honour him, here's a snippet using his code to find the first or all sums with a certain value in an Array

 const log = (...strs) => document.querySelector("pre").textContent += `${strs.join(" ")}\\n`; const array = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5]; log("All possible sums 28 in array:", findSumsInArray(array, 28)); log("First sum 28 in array:", findSumsInArray(array, 28, true)); log("All possible sums 22 in array:", findSumsInArray(array, 22)); log("First sum 22 in array:", findSumsInArray(array, 22, true)); log("All possible sums 12 in array:", findSumsInArray(array, 12)); log("First sum 12 in array:", findSumsInArray(array, 12, true)); log("First sum 28 in array [3, 5, 7, 14, 14]:", findSumsInArray([3, 5, 7, 14, 14], 28, true)); function findSumsInArray(array, sum2Find, firstOnly) { let set = new Set(array); let result = []; const errorStr = `Sum ${sum2Find} not possible in [${array}]`; for (let item of array) { let num = sum2Find - item; if (set.has(num)) { const report = `${num}+${item}`; if (firstOnly) { return report; } result.push(report); set.delete(item); } } return result.length ? `[${result.join(", ")}]` : errorStr; } 

 <pre></pre> 

Answer 11

I'm a bit late in contributing, but no one appears to have posted the obligatory JavaScript one-liner.

let input = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5];
let dSum = 28;
let Result = [];

input.forEach( i=>{ input.forEach( j=>{ i+j==dSum && Result.push([i,j]); }); });

The Result array will contain the solutions, of which there are two in this case:

[[12, 16], [16, 12]]

Generalising this to a function where values representing input and dSum are user-defined parameters:

pairsFromArrayWithSum = (total, Numbers) 
                      => Numbers.forEach(a => { 
                         Numbers.forEach(b => { 
                         a + b == total && Result.push([a,b]);
                         }); });

Then, given:

Result = [];
pairsFromArrayWithSum(50, [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]);

the Result array will contain:

[ [ 3, 47 ],
  [ 7, 43 ],
  [ 13, 37 ],
  [ 19, 31 ],
  [ 31, 19 ],
  [ 37, 13 ],
  [ 43, 7 ],
  [ 47, 3 ] ]

Answer 12

 let input = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5]; let dSum = 28; let set = new Set(); for(item of input) { let num = dSum - item; if(set.has(num)) { console.log(num + " " + item); break; } set.add(item); }

 function detectPair(sum, array) { for (i=0; i<array.length; i++) { for (j=0; j<array.length; j++) { if (i == j) continue; else if (array[i] + array[j] === sum) return [array[i], array[j]]; } }; return null; } let sum = 28; let array = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5] console.log(detectPair(sum, array)); //return would be [12,16]

 function detectPair(sum, array) { for (i=0; i<array.length; i++) { for (j=0; j<array.length; j++) { if (i == j) continue; else if (array[i] + array[j] === sum) return [array[i], array[j]]; } }; return null; } let sum = 28; let array = [3, 5, 7, 9, 4, 8, 5, 12, 4, 9, 16, 5] console.log(detectPair(sum, array)); //return would be [12,16]