I'm solving basic dynamic programming question. It's simple. input is single integer. if input is 3, all possible cases is 3,1+2,2+1,1+1+1 so out put is 4. if input is 4, output is 7 because 4 = 1+1+1+1 1+1+2 1+2+1 2+1+1 2+2 1+3 3+1. code above is my code to solve this. 

use strict;
use warnings;

our @cache;
sub find {
    my $num = shift;
    if(defined $cache[$num]) {return $cache[$num];}
    if($num == 0) {return 1;}
    if($num < 0) {return 0;}
    
    $cache[$num] = &find($num -1) + &find($num - 2) + &find($num - 3);
    return $cache[$num];
}

my $test_case = <>+0;
for(1..$test_case) {
    my $num = <>+0;
    print &find($num),"\n";
}

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but the output is wrong. and I found that output of 2 is 3, instead of 2. also output of 3 is 5 instead of 4. to fix this, I write @cache1..3 = (1,2,4); and of course, It is fixed. But, I want to remove @cache1..3 = (1,2,4); and want to know why my original code is not working properly. when call solution(2), isn't it obvious that the output is solution(1) + solution(0) + solution(-1)? and when I call solution(0), solution(1), solution(-1) individually, the output is 1,1,0 so sum is 2. not 3. Thank you for your help.

In reply to problem while solving basic dynamic programming question by yujong_lee

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