Meet In The Middle Technique | Leetcode Partition Array Into Two Arrays to Minimize Sum Difference
Meet In The Middle Technique | Leetcode Partition Array Into Two Arrays to Minimize Sum Difference | CSES Problem set
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Пікірлер: 27
nice video very helpful
was initially trying to solve this problem using memoization but it was showing memory limit exceeded but this approach was better, thanks anmol bhaiya :)
@happyengineeringwithanmol
5 ай бұрын
Thanks man
This is really good and thorough! Kudos Anmol!! Keep Going!
@happyengineeringwithanmol
Жыл бұрын
Thanks man
Love you, thank you for the detail explanation.
@happyengineeringwithanmol
4 ай бұрын
Thanks
thanks bro great explanation but can u explain the lower bound part in a bit depth why b should be >=(sum-2*a)/2;
nice video understood nicely :)
@happyengineeringwithanmol
2 жыл бұрын
Thanks man
Bhai kya he sahi socha h Awesome !! kitna practice kiya ho bhai itna soch lane ke liye ?
@happyengineeringwithanmol
Жыл бұрын
Bhai tejj hi hoo bachpan sa mein😂😂 jokes apart I have been doing dsa for past 4 years.
can you explain sum - 2*(1 + sum2)
in the formula sum-2*(sum1) , why u wrote 1+sum2
@iloveuniverse143
11 ай бұрын
same doubt , in example it should be 22 - 2*(3 + 7) = 2
did you solve the whole cses problem sheet?
@happyengineeringwithanmol
5 ай бұрын
Tried to solve the maximum I could
Which software are you using to write?
@happyengineeringwithanmol
11 ай бұрын
Obs
@happyengineeringwithanmol
11 ай бұрын
Good notes
full confusion bro
@happyengineeringwithanmol
Жыл бұрын
Watch it again
here is a solution of the leetcode problem class Solution { public int minimumDifference(int[] nums) { if(nums.length == 2) return Math.abs(nums[0] - nums[1]); int total = Arrays.stream(nums).sum(); int n = nums.length / 2; HashMap map1 = new HashMap(); HashMap map2 = new HashMap(); for(int i = 0 ; i if((bitmask&1) == 1){ sumbits++; sum += nums[pos]; } pos++; bitmask = bitmask >> 1; } if(!map1.containsKey(sumbits)) map1.put(sumbits,new ArrayList()); map1.get(sumbits).add(sum); } for(int i = 0 ; i if((bitmask&1) == 1){ sumbits++; sum += nums[pos + n]; } pos++; bitmask = bitmask >> 1; } if(!map2.containsKey(sumbits)) map2.put(sumbits,new ArrayList()); map2.get(sumbits).add(sum); } for(int i = 0 ; i
elements can be negative also, so ig your solution is wrong.
@happyengineeringwithanmol
2 жыл бұрын
The solution is correct its accepted on both the platforms I feel you missed out somewhere you can join my telegram group for detailed discussion
@adityaagarwal1650
2 жыл бұрын
@@happyengineeringwithanmol i mean in the explanation it be abs(20 - 2*sum2) >= 0 and one doubt I have is auto itr = lower_bound(v.begin(), v.end(),b) for this we should also consider one iterator prior to itr also (i.e. (-- itr)) as that also give us minimum?.. right?
@happyengineeringwithanmol
2 жыл бұрын
@@adityaagarwal1650 even I did added that condition previously but test cases were like that my solution passed