Hi there – In this challenge I’m wondering if the problem spec and one of the test cases are in conflict. I’m having problems specifically with this test case:

`pairwise([0, 0, 0, 0, 1, 1], 1)`

My code outputs “4” for this test, but the directions say it should be “10”, which I assume is because it is adding the index pairs (0,4) and (1,5) for 0+4+1+5 = 10. (Indices 0 and 4 correspond to element values 0 and 1, which add up to 1, while indices 1 and 5 correspond to element values 0 and 1, which add up to 1.)

However, the directions say “If multiple pairs are possible that have the same numeric elements but different indices, return the smallest sum of indices.” For that test case, the index pair (0,4) has element values of 0 and 1, while the index pair (1,5) also has element values of 0 and 1. Those are the same numeric elements, right? So shouldn’t the program only output the smallest sum of 0+4 while ignoring 1+5?

I don’t understand how 10 is the correct output, unless the spec really means to say: “If multiple pairs are possible that have the same numeric elements and also one shared index, then return the smallest sum of indices; but otherwise if the numeric elements are the same and both index pairs are different, then add both index pairs.” All of that is really oddly specific, so I don’t think that was the intention. Any help understanding this?

Thanks for the guidance. I’m pretty sure my code is more complicated than it has to be to do this regardless of that test case, but any tips are appreciated!

**Your code so far**

```
function pairwise(arr, arg) {
var out;
// find element pairs in arr that sum to arg
var pairelems = [];
var pairindices = [];
var pairsum = [];
// search through arr for pairs; don't reuse indices
for(var i=0; i<arr.length; i++){
for(var j=i+1; j<arr.length; j++){
if(arr[i]+arr[j]===arg){
// save numeric pairs; put smallest first for easy compare later
var small;
var large;
if(arr[i]<arr[j]){
small = arr[i];
large = arr[j];
}
else{
small = arr[j];
large = arr[i];
}
pairelems.push([small,large]);
pairindices.push([i,j]);
// save sum of indices of pairs
pairsum = pairsum.concat(i+j);
}
}
}
// if any repeat pairs, note lowest sum of indices
var indexRemove = [];
for(var k=0; k<pairelems.length; k++){
for(var l=k+1; l<pairelems.length; l++){
// check if any repeat element pairs
if(pairelems[k][0]===pairelems[l][0] &&
pairelems[k][1]===pairelems[l][1]){
// save indices of repeats if not already saved once
if(indexRemove.indexOf(k)<0){
indexRemove.push(k);
}
if(indexRemove.indexOf(l)<0){
indexRemove.push(l);
}
}
}
}
//remove indices denoted EXCEPT the one for the smallest pairsum
var ordered = pairsum;
ordered.sort(function(a,b){
return a-b;
});
indexRemove.splice(pairsum.indexOf(ordered[0]),1);
for(var m=0; m<indexRemove.length; m++){
pairsum[indexRemove[m]]=0; // set to zero so it won't add up
}
// output sum of all sums of indices
out = pairsum.reduce(function(a,b){
return a+b;
});
//out = pairelems; //debug
// out = pairsum; //debug
return out;
}
//pairwise([1,4,2,3,0,5], 7);
//pairwise([1, 1, 1], 2);
pairwise([0, 0, 0, 0, 1, 1], 1);```
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**Link to the challenge:**
https://www.freecodecamp.org/challenges/pairwise
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