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1137. N-th Tribonacci Number (Iterative)

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	`1`	`+package com.cheehwatang.leetcode;`
	`2`	`+`
	`3`	`+// Time Complexity : O(n),`
	`4`	`+// where 'n' is the input 'n'.`
	`5`	`+// We traverse from 3 to 'n' to calculate the n-th tribonacci number.`
	`6`	`+//`
	`7`	`+// Space Complexity : O(1),`
	`8`	`+// as the auxiliary space used is independent of the input.`
	`9`	`+`
	`10`	`+public class NthTribonacciNumber_Iterative {`
	`11`	`+`
	`12`	`+ // Approach:`
	`13`	`+ // We use the iterative method, using the variable 'first', 'second' and 'third' to keep track.`
	`14`	`+`
	`15`	`+ public int tribonacci(int n) {`
	`16`	`+ if (n == 0 \|\| n == 1) return n;`
	`17`	`+ if (n == 2) return 1;`
	`18`	`+`
	`19`	`+ // As we need both 'n - 1', 'n - 2' and 'n - 3' to start, record 'first', 'second' and 'third', respectively.`
	`20`	`+ // For each iteration until n, sum all three numbers and replace the numbers.`
	`21`	`+ int first = 1;`
	`22`	`+ int second = 1;`
	`23`	`+ int third = 0;`
	`24`	`+ int result = 0;`
	`25`	`+ for (int i = 3; i <= n; i++) {`
	`26`	`+ result = first + second + third;`
	`27`	`+ third = second;`
	`28`	`+ second = first;`
	`29`	`+ first = result;`
	`30`	`+ }`
	`31`	`+ return result;`
	`32`	`+ }`
	`33`	`+}`

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