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Created September 23, 2016 17:12
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JS Bin [add your bin description] // source https://jsbin.com/tulemun
Raw
index.html
<!DOCTYPE html>
<html>
<head>
<meta name="description" content="[add your bin description]">
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JS Bin</title>
</head>
<body>
<script id="jsbin-javascript">
/**
This is a solution for a generalization of a problem i was asked in an intereview.
The first question was for a given array of strings to print all possible pairs where order of pairs is not important.
for example, for a given array ["a","b","c"] the output shoult be :
"a","b",
"a","c",
"b","c"
I solve this queestion in atrivial way but two nested loops.
The the interviewer asked me how would i solve this for a more general question
so we have N combinaiton instead of 2 in the given original question.
I didn't manage to solve that in the interview but i picked my brain and 3 hours of the interview
i scatch the following solution on a piece of paper in the airport while waiting for a relative.
So now that i solved it, i can finally give my mind some rest.
i hope the solution is readable and understandable, although it is a bit complex.
*/
"use strict";
var arr = ["a", "b", "c", "d", "e"];
//this function is pretty simple, it just console log all the
//current combination based on an array of indexers which tells me where we are.
function pairs(arr, n) {
//initializing the indexers array.
var indexers = [];
for (var i = 0; i < n; i++) {
indexers.push(i);
}
var stop = false;
while (!stop) {
console.log(indexers.map(function (index) {
return arr[index];
}));
stop = moveIndexers(indexers, arr.length);
}
}
//this is the main function which more the indexers.
//it returns true in case there is no more options to run over.
function moveIndexers(indexers, len) {
var lastIndex = indexers.length - 1;
var current = lastIndex;
while (current >= 0) {
if (indexers[current] < len - 1 - (lastIndex - current)) {
for (var i = current; i <= lastIndex; i++) {
if (i === current) {
indexers[i]++;
} else {
indexers[i] = indexers[i - 1] + 1;
}
}
return;
}
current--;
if (current < 0) return true;
}
}
pairs(arr, 2);
pairs(arr, 3);
</script>
<script id="jsbin-source-javascript" type="text/javascript">/**
This is a solution for a generalization of a problem i was asked in an intereview.
The first question was for a given array of strings to print all possible pairs where order of pairs is not important.
for example, for a given array ["a","b","c"] the output shoult be :
"a","b",
"a","c",
"b","c"
I solve this queestion in atrivial way but two nested loops.
The the interviewer asked me how would i solve this for a more general question
so we have N combinaiton instead of 2 in the given original question.
I didn't manage to solve that in the interview but i picked my brain and 3 hours of the interview
i scatch the following solution on a piece of paper in the airport while waiting for a relative.
So now that i solved it, i can finally give my mind some rest.
i hope the solution is readable and understandable, although it is a bit complex.
*/
var arr = ["a","b","c","d","e"];
//this function is pretty simple, it just console log all the
//current combination based on an array of indexers which tells me where we are.
function pairs(arr, n){
//initializing the indexers array.
var indexers = [];
for( var i=0; i<n; i++ ){
indexers.push(i);
}
var stop = false;
while( !stop ){
console.log(indexers.map((index) => arr[index]));
stop = moveIndexers(indexers, arr.length);
}
}
//this is the main function which more the indexers.
//it returns true in case there is no more options to run over.
function moveIndexers(indexers, len){
var lastIndex = indexers.length-1;
var current = lastIndex;
while( current >= 0 ){
if( indexers[current] < (len-1)-(lastIndex-current)){
for( var i=current; i<=lastIndex; i++ ){
if( i=== current ){
indexers[i]++;
}
else{
indexers[i] = indexers[i-1]+1;
}
}
return;
}
current--;
if( current < 0 ) return true;
}
}
pairs(arr,2)
pairs(arr,3);</script></body>
</html>
Raw
jsbin.tulemun.js
/**
This is a solution for a generalization of a problem i was asked in an intereview.
The first question was for a given array of strings to print all possible pairs where order of pairs is not important.
for example, for a given array ["a","b","c"] the output shoult be :
"a","b",
"a","c",
"b","c"
I solve this queestion in atrivial way but two nested loops.
The the interviewer asked me how would i solve this for a more general question
so we have N combinaiton instead of 2 in the given original question.
I didn't manage to solve that in the interview but i picked my brain and 3 hours of the interview
i scatch the following solution on a piece of paper in the airport while waiting for a relative.
So now that i solved it, i can finally give my mind some rest.
i hope the solution is readable and understandable, although it is a bit complex.
*/
"use strict";
var arr = ["a", "b", "c", "d", "e"];
//this function is pretty simple, it just console log all the
//current combination based on an array of indexers which tells me where we are.
function pairs(arr, n) {
//initializing the indexers array.
var indexers = [];
for (var i = 0; i < n; i++) {
indexers.push(i);
}
var stop = false;
while (!stop) {
console.log(indexers.map(function (index) {
return arr[index];
}));
stop = moveIndexers(indexers, arr.length);
}
}
//this is the main function which more the indexers.
//it returns true in case there is no more options to run over.
function moveIndexers(indexers, len) {
var lastIndex = indexers.length - 1;
var current = lastIndex;
while (current >= 0) {
if (indexers[current] < len - 1 - (lastIndex - current)) {
for (var i = current; i <= lastIndex; i++) {
if (i === current) {
indexers[i]++;
} else {
indexers[i] = indexers[i - 1] + 1;
}
}
return;
}
current--;
if (current < 0) return true;
}
}
pairs(arr, 2);
pairs(arr, 3);
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