solve coding problems data structures and algorithm problems

YOU CANNOT USE ANY BUILT$-$IN FUNCTION EXCEPT len$$IN PYTHON.

$1.$ Merge Lineup:

Now you guys are playing a $2$v$2$ fierce pokemon battle. At one point your and your teammate's pokemons have very low hp$.$ So you created a new rule to merge the hp of your and your teammate's pokemons and create a new pokemon lineup. But the opposite team placed a special condition. You and your teammate have to add the hp of your team's pokemons from opposite directions. That is$,$ your first pokemon's hp will be added to your teammate's last pokemon's hp; your second pokemon's hp will be added to your teammate's second last pokemon's hp and so on and so forth. The None elements denote $0$ hp$.$ You and your teammate have the same number of pokemons. Implement the scenario using a proper method which takes your and your teammate's pokemon hp as two arrays $($as parameters$)$; and returns the resulting arr.

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Sample Input $/$ Driver Code:

pokemon$\_1$: $[4,5,-1,$ None, None$]$

pokemon$\_2$: $[2,27,7,12,$ None$]$

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#Function Call:$$ print$($mergeLineup$($pokemon$\_1,$ pokemon$\_2))$

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Sample Output:

$[4,17,6,27,2]$

$$#drivercode

def mergeLineup$($pokemon$\_1,$ pokemon$\_2)$:

$$ #To Do

print$("///$Task $01$: Merge Lineup $///")$

pokemon$\_1=$ np$.$array$([12,3,25,1,$ None$])$

pokemon$\_2=$ np$.$array$([5,-9,3,$ None, None$])$

returned$\_$value $=$mergeLineup$($pokemon$\_1,$ pokemon$\_2)$

print$($f$\text{'}$Task $1$: $\{$returned$\_$value$\}\text{'})$ # This should print $[12,3,28,-8,5]$

unittest.output$\_$test$($returned$\_$value, np$.$array$([12,3,28,-8,5]))$

pokemon$\_1=$ np$.$array$([4,5,-1,$ None, None$])$

pokemon$\_2=$ np$.$array$([2,27,7,12,$ None$])$

returned$\_$value $=$mergeLineup$($pokemon$\_1,$ pokemon$\_2)$

print$($f$\text{'}$Task $1$: $\{$returned$\_$value$\}\text{'})$ # This should print $[4,17,6,27,2]$

unittest.output$\_$test$($returned$\_$value, np$.$array$([4,17,6,27,2]))$

Explanation:

$4+$None$(0)=4\{$pokemon$\_1[0]+$pokemon$\_2[4]\},$

$5+12=17\{$pokemon$\_1[1]+$pokemon$\_2[3]\},$

$-1+7=6\{$pokemon$\_1[2]+$pokemon$\_2[2]\},$

None$(0)+27=27\{$pokemon$\_1[3]+$pokemon$\_2[1]\},$

None$(0)+2=2\{$pokemon$\_1[4]+$pokemon$\_2[0]\}$

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