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refactor: Enhance docs, code, add tests in LucasSeries (#6749)
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@@ -1,38 +1,69 @@
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package com.thealgorithms.maths;
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/**
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* https://en.wikipedia.org/wiki/Lucas_number
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* Utility class for calculating Lucas numbers.
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* The Lucas sequence is similar to the Fibonacci sequence but starts with 2 and
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* 1.
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* The sequence follows: L(n) = L(n-1) + L(n-2)
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* Starting values: L(1) = 2, L(2) = 1
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* Sequence: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, ...
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*
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* @see <a href="https://en.wikipedia.org/wiki/Lucas_number">Lucas Number</a>
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* @author TheAlgorithms Contributors
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*/
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public final class LucasSeries {
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private LucasSeries() {
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}
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/**
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* Calculate nth number of Lucas Series(2, 1, 3, 4, 7, 11, 18, 29, 47, 76,
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* 123, ....) using recursion
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* Calculate the nth Lucas number using recursion.
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* Time Complexity: O(2^n) - exponential due to recursive calls
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* Space Complexity: O(n) - recursion depth
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*
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* @param n nth
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* @return nth number of Lucas Series
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* @param n the position in the Lucas sequence (1-indexed, must be positive)
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* @return the nth Lucas number
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* @throws IllegalArgumentException if n is less than 1
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*/
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public static int lucasSeries(int n) {
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return n == 1 ? 2 : n == 2 ? 1 : lucasSeries(n - 1) + lucasSeries(n - 2);
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if (n < 1) {
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throw new IllegalArgumentException("Input must be a positive integer. Provided: " + n);
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}
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if (n == 1) {
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return 2;
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}
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if (n == 2) {
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return 1;
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}
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return lucasSeries(n - 1) + lucasSeries(n - 2);
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}
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/**
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* Calculate nth number of Lucas Series(2, 1, 3, 4, 7, 11, 18, 29, 47, 76,
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* 123, ....) using iteration
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* Calculate the nth Lucas number using iteration.
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* Time Complexity: O(n) - single loop through n iterations
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* Space Complexity: O(1) - constant space usage
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*
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* @param n nth
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* @return nth number of lucas series
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* @param n the position in the Lucas sequence (1-indexed, must be positive)
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* @return the nth Lucas number
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* @throws IllegalArgumentException if n is less than 1
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*/
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public static int lucasSeriesIteration(int n) {
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if (n < 1) {
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throw new IllegalArgumentException("Input must be a positive integer. Provided: " + n);
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}
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if (n == 1) {
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return 2;
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}
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if (n == 2) {
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return 1;
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}
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int previous = 2;
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int current = 1;
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for (int i = 1; i < n; i++) {
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for (int i = 2; i < n; i++) {
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int next = previous + current;
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previous = current;
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current = next;
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}
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return previous;
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return current;
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}
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}
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