help me solve this and for test cases I need all Possible ones URGENT!! Smell$1$Almostbest.java$-$ package org.example;

import java.util.HashMap;

import java.util.Map;

$//$ This code will return the $1$st argument to the power of all subsequent arguments

$//$ This one caches our results for later reuse

public class Smell$1$AlmostBest $\{$

private static Map$>\_\_$cache $=$ new HashMap$>()$;

public static void main$($String$[]$ args$)\{$

if $($args$.$length $<mtext></mtext><mn>2</mn><mi>)</mi><mtext></mtext><mi>\{</mi>$

System.out.println$("$Usage: java ser$515.$smells.Smell$1$AlmostBest $+")$;

System.exit$(-1)$;

$\}$

int num $=$ Integer.parseInt$($args$[0])$;

long timer $=$ System.currentTimeMillis$()$;

for $($int j $=1$; j $$ args.length; j$++)\{$

System.out.println$($

"The $"+$ args$[$j$]+"$th power of $"+$ num $+"$ is $"+$ toPower$($num$,$ Integer.parseInt$($args$[$j$])))$;

$\}$

System.out.println$("$Completed $"+$ args$[1]+"$ iterations in $"+($System$.$currentTimeMillis$()-$ timer$)+"$ms$")$;

$\}$

$//$ What is the inefficiency here?

public static int toPower$($int n$,$ int pow$)\{$

Map entry $=\_\_$cache.get$($n$)$;

Integer res $=$ null;

if $($entry $!=$ null$)\{$

res $=$ entry.get$($pow$)$;

if $($res $==$ null$)\{$

res $=$ calcPower$($n$,$ pow$)$;

entry.put$($pow$,$ res$)$;

$\}$

$\}$ else $\{$

res $=$ calcPower$($n$,$ pow$)$;

entry $=$ new HashMap$()$;

entry.put$($pow$,$ res$)$;

$\_\_$cache.put$($n$,$ entry$)$;

$\}$

return res;

$\}$

private static int calcPower$($int n$,$ int pow$)\{$

if $($pow $==0)$

return $1$;

int res $=1$;

for $($int i $=0$; i $$ pow; res $*=$ n$,$ i$++)$

;

;

;

return res;

$\}$

$\}.$

IPower.java$-$ package org.example;

import java.util.HashMap;

import java.util.Map;

public interface IPower $\{$

int toPower$($int n$,$ int pow$)$;

$\}$

class PowerSimple implements IPower $\{$

public int toPower$($int n$,$ int pow$)\{$

if $($pow $==0)$

return $1$;

int res $=1$;

for $($int i $=0$; i $$ pow; res $*=$ n$,$ i$++)$

;

;

;

return res;

$\}$

$\}$

class PowerCached implements IPower $\{$

private static Map$>\_\_$cache $=$ new HashMap$>()$;

$//$ resolves the inefficiency in AlmostBest

public int toPower$($int n$,$ int pow$)\{$

Map entry $=$ PowerCached.$\_\_$cache.get$($n$)$;

if $($entry $==$ null$)\{$

entry $=$ new HashMap$()$;

$\}$

return toCachedPower$($entry$,$ n$,$ pow$)$;

$\}$

$//$ The contract here is we know there is a cache entry so no check required

$//($DBC$)$

private int toCachedPower$($Map e$,$ int n$,$ int pow$)\{$

Integer res $=$ null;

if $($pow $==0)\{$

res $=1$;

$\}$ else $\{$

res $=$ e$.$get$($pow$)$;

if $($res $!=$ null$)\{$

return res;

$\}$ else $\{$

res $=$ n $*$ toCachedPower$($e$,$ n$,$ pow $-1)$;

e$.$put$($pow$,$ res$)$; $//$ saves each intermediate result in the cache

$\}$

$\}$

return res;

$\}$

$\}$