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1207-UniqueNumOccur.ts
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1207-UniqueNumOccur.ts
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/*
Given an array of integers arr, return true if the number of
occurrences of each value in the array is unique or false otherwise.
Solution: Keep track of the number of times a value comes up;
then compare the lengths of the resultant Object and the Set of Object values.
If the lengths match, then there is a unique number of occurrences, because
every value for every key would be unique. If it were false, then 2 or
more values would be the same, and the Set of values would be shorter than
the length of the hash Object.
Runtime Complexity: O(n) We iterate over the array and constant-time insert into the Object
Space Complexity: O(n) We're storing the length of the hashValues
*/
function uniqueOccurrences(arr: number[]): boolean {
const hash = {};
for (let i = 0; i < arr.length; i += 1) {
if (arr[i] in hash) {
hash[arr[i]] += 1;
} else {
hash[arr[i]] = 1;
}
}
const hashValues = Object.values(hash);
return hashValues.length === new Set(hashValues).size;
};