Merge Sorted Array | LeetCode 88 | Coding Interview Tutorial

Merge Sorted Array solution: LeetCode 88
Code and written explanation: terriblewhiteboard.com/merge-...
Link to problem on LeetCode: leetcode.com/problems/merge-s...
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Merge Sorted Array | LeetCode 88 | Coding Interview
#mergesortedarray #leetcode #algorithms #terriblewhiteboard #codinginterview
Click the time stamp to jump to different parts of the video.
00:00 Title
00:06 Problem readout
00:56 Whiteboard solution
03:40 Coding solution
10:40 Result and outro

[@TerribleWhiteboard]

This solution is valid for JavaScript. You may have to modify it to fit your particular language.
Java solution:
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int first = m - 1;
int second = n - 1;
for (int i = nums1.length - 1; i >= 0; i--) {
if (second < 0) {
return;
}
if (first >= 0 && nums1[first] > nums2[second]) {
nums1[i] = nums1[first];
first--;
} else {
nums1[i] = nums2[second];
second--;
}
}
}
}

[@epk1001]

I got stuck at the test case [0], 0, [1],1, I missed the first >= 0 condition. Thanks for your java solution. - :)

[@quangngo7264]

thanks for your solution

[@YKLUO]

such a beautiful solution

[@TerribleWhiteboard]

Thanks!

[@aigulzhyldyzbekova3480]

I totally agree

[@rahulsbhatt]

It's a very good solution, thank you so much!

[@TerribleWhiteboard]

Thanks!

[@15kfelix]

need to add:
if(first < 0){
nums1[i] = nums2[second];
second--;
}
to account for new test cases

[@ibrahimalkuwaifi6087]

Actually you need to add this
if first < 0:
while second >= 0:
nums1[replaceIndex] = nums2[second]
second-=1
replaceIndex-=1

[@cocoarecords]

wow thanks

[@TerribleWhiteboard]

You're welcome!

[@AbhishekKumar-ff9vg]

good work

[@adarshmv261]

can you put the python version of this pls?

[@TerribleWhiteboard]

I would, but I don't have time to make videos in more than one language. I'm hoping the whiteboarding can be translated to multiple languages, though.

[@louistannudin2486]

leetcode.com/problems/merge-sorted-array/discuss/29522/This-is-my-AC-code-may-help-you

[@nhingo6067]

Any idea why the solution fails if I switch the order of if conditions as: if (nums1[first]

[@leecharlie2513]

Why start from the end? And what is the difference if starting from front?

[@derekmurray5974]

Hello, I am wondering this as well. Please advise. Thanks a lot!

[@derekmurray5974]

Hi, I'm revisiting this to see why start from the end and whats the difference....Thanks a lot for your contribution through creating this page

[@SuhasSrivats]

It fails for this test-case:
Input:
[2,0]
1
[1]
1
Output:
[2,2]
Expected:
[1,2]

[@louistannudin2486]

leetcode.com/problems/merge-sorted-array/discuss/29522/This-is-my-AC-code-may-help-you

[@15kfelix]

add if(first < 0){
nums1[i] = nums2[second];
second--;
}
to top of if/else chain

[@singforhappiness1]

first = m - 1
second = n - 1
for i in range (m+n-1,-1,-1):
if second < 0 :
break;
elif first < 0:
nums1[i] = nums2[second]
second -= 1
elif nums1[first] > nums2[second]:
nums1[i] = nums1[first]
first -= 1
else:
nums1[i] = nums2[second]
second -= 1

[@vaishalirockzz]

This solution is only valid if m>n.
it will not work if m

[@TerribleWhiteboard]

It works for javascript. You might have to make minor adjustments if you're using another language.

[@KaranKumar-xt2mh]

It's literally given in the question to merge nums2 into nums1

[@mandeepashiya2229]

A note from the question - You may assume that nums1 has enough space (size that is equal to m + n) to hold additional elements from nums2.

[@tapanpanchal757]

I can't pass one test case i.e:
[0]
0
[1]
1
since, first = m-1, and m is 0 in this case, so first becomes negative and it gives a runtime error
hope you see this

[@shadmanmartinpiyal4057]

same here! any solution u found ?

[@jhessikanda]

Solution covering edge cases mentioned here:
var merge = function(nums1, m, nums2, n) {
let firstLastElementIndex = m -1;
let secondLastElementIndex = n -1;
for (let x=(m + n -1); x >= 0; x--) {
if (secondLastElementIndex < 0) {
break;
}
if (nums1[firstLastElementIndex] < nums2[secondLastElementIndex] ||
firstLastElementIndex < 0 ||
m === 0) {
nums1[x] = nums2[secondLastElementIndex];
secondLastElementIndex--;
} else {
nums1[x] = nums1[firstLastElementIndex];
firstLastElementIndex--;
}
}
};

[@ashutoshmaheshwari]

you missed a condition..what if first

[@hideinbush0]

Why is there no return statement to show the output?

[@real_chimps9570]

Because you're supposed to modify nums1, you don't need to return anything to do that

[@AhmedHemaz]

I am dump that's it