Is there a way to determine if two sets of integers in JavaScript are permutations? For example, given the arrays:
a = [1, 2, 3, 4, 5]
and
b = [2, 3, 5, 1, 4]
I want a function that will return true if they are permutations of each other.
Is there a way to determine if two sets of integers in JavaScript are permutations? For example, given the arrays:
a = [1, 2, 3, 4, 5]
and
b = [2, 3, 5, 1, 4]
I want a function that will return true if they are permutations of each other.
To keep track of occurrences, consider using a Map. Update the count in the map whenever you encounter a matching element in the second array:
function checkPermutation(arr1, arr2) {
if (arr1.length !== arr2.length) return false;
let occurrenceMap = new Map(arr1.map(item => [item, { count: 0 }]));
arr1.forEach(item => occurrenceMap.get(item).count++);
return arr2.every(item => {
let match = occurrenceMap.get(item);
return match && match.count--;
});
}
let arr1 =[1,2,3,4,5,1];
let arr2 = [2,3,1,5,1,4];
console.log(checkPermutation(arr1, arr2));A simple solution that requires minimal effort:
let x = [5,4,3,2,1],
y = [4,2,3,1,5];
let result = JSON.stringify(x.sort()) === JSON.stringify(y.sort())
console.log(result)An optimized approach:
function isPermutation (x,y) {
let map = x.reduce((acc,c) => {acc[c] = (acc[c] || 0) + 1; return acc},{})
for (element of y) {
if (!map[element] || map[element] == 0) {
return false;
} else {
map[element]--;
}
}
for (key in map) {
if (map[key] != 0) {
return false;
}
}
return true;
}
console.log(isPermutation([5, 4, 3, 2, 1],[4, 2, 3, 1, 5])) // => true
console.log(isPermutation([1, 2, 3],[3, 2, 2])) // => false
console.log(isPermutation([1,4,5],[5,1,4])) // => trueThis method bears similarities to another solution, but it provides a slightly different perspective.
The logic behind this approach involves creating a map from the first array using reduce, where the value represents the count of occurrences. The second array's elements are then used to subtract counts from the corresponding keys in the map. If a key doesn't exist or its value is already zero, the arrays are not permutations. Finally, the program checks if all values in the map are precisely zero.
let num1 = [1, 2, 3, 4, 5];
let num2 = [2, 3, 5, 1, 4];
return num1.filter(number => !num2.includes(number)).length === 0
This code snippet will yield true if all values in num1 are present in num2, regardless of their order.
Success:
let x = [7,8,9,2,6,5];
let y = [8,9,5,2,6,7];
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