The task is to get the Cartesian product of the arrays (Finding all combinations after concatenating them). Here, 2 approaches are discussed with the help of JavaScript.
Approaches:
- using JavaScript Array.reduce() method
- using recursive approach
Approach 1:
- First, declare an array of multiple arrays.
- Recursion is used to solve the problem. The base condition is when the length of the array reduces to zero then return the string build till now. Else
- Reduce the first array element by reduce() method and return the result returned from the recursion result (Recursion is called every time after leaving the first item of the array) plus the previous value in concatenation with every array element.
- Return the final ans array, which contains all combinations.
Example 1: This example implements the above approach using JavaScript Aarry.reduce() method
Javascript
let arr = [ [ 'm' , 'n' ],
[ 'c' ],
[ 'd' , 'e' , 'f' ]
]; function getCombn(arr, pre) {
pre = pre || '' ;
if (!arr.length) {
return pre;
}
let ans = arr[0].reduce( function (ans, value) {
return ans.concat(getCombn(
arr.slice(1), pre + value));
}, []);
return ans;
} console.log(getCombn(arr)); |
Output
[ 'mcd', 'mce', 'mcf', 'ncd', 'nce', 'ncf' ]
Approach 2:
- First, declare an array of multiple arrays.
- Recursion is used to solve the problem. The base condition is When the length of the array reduces to one then return that element of the array. Else
- Call recursion after leaving the first element of the array and store the result in a variable(otherCases).
- Loop through every element of the Array (otherCases) and inside every element loop through the first element of the Array(arr).
- Concatenate every element of the Array(arr[0]) with the Array(otherCases) and push the result in the answer array.
Example 2: This example implements the above approach using recursion in JavaScript.
Javascript
// Create new array let arr = [ [ 'm' , 'n' ],
[ 'c' ],
[ 'd' , 'e' , 'f' ]
]; // Funtion to get all combinations function getCombn(arr) {
// When array contain 1 element
if (arr.length == 1) {
return arr[0];
} else {
let ans = [];
// Recur with the rest of the array.
let otherCases = getCombn(arr.slice(1));
for (let i = 0; i < otherCases.length; i++) {
for (let j = 0; j < arr[0].length; j++) {
ans.push(arr[0][j] + otherCases[i]);
}
}
return ans;
}
} // Display output console.log(getCombn(arr)); |
Output
[ 'mcd', 'ncd', 'mce', 'nce', 'mcf', 'ncf' ]