We\'re given a string and a permutation of the string.
For example, an input string sandeep and a permutation psdenae.
Find the pos
Building off @Algorithmist 's answer, and his comment to his answer, and using the principle discussed in this post for when there are repeated letters, I made the following algorithm in JavaScript that works for all letter-based words even with repeated letter instances.
function anagramPosition(string) {
var index = 1;
var remainingLetters = string.length - 1;
var frequencies = {};
var splitString = string.split("");
var sortedStringLetters = string.split("").sort();
sortedStringLetters.forEach(function(val, i) {
if (!frequencies[val]) {
frequencies[val] = 1;
} else {
frequencies[val]++;
}
})
function factorial(coefficient) {
var temp = coefficient;
var permutations = coefficient;
while (temp-- > 2) {
permutations *= temp;
}
return permutations;
}
function getSubPermutations(object, currentLetter) {
object[currentLetter]--;
var denominator = 1;
for (var key in object) {
var subPermutations = factorial(object[key]);
subPermutations !== 0 ? denominator *= subPermutations : null;
}
object[currentLetter]++;
return denominator;
}
var splitStringIndex = 0;
while (sortedStringLetters.length) {
for (var i = 0; i < sortedStringLetters.length; i++) {
if (sortedStringLetters[i] !== splitString[splitStringIndex]) {
if (sortedStringLetters[i] !== sortedStringLetters[i+1]) {
var permutations = factorial(remainingLetters);
index += permutations / getSubPermutations(frequencies, sortedStringLetters[i]);
} else {
continue;
}
} else {
splitStringIndex++;
frequencies[sortedStringLetters[i]]--;
sortedStringLetters.splice(i, 1);
remainingLetters--;
break;
}
}
}
return index;
}
anagramPosition("ARCTIC") // => 42
I didn't comment the code but I did try to make the variable names as explanatory as possible. If you run it through a debugger process using your dev tools console and throw in a few console.logs you should be able to see how it uses the formula in the above-linked S.O. post.