Meet In The Middle Technique | Leetcode Partition Array Into Two Arrays to Minimize Sum Difference
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- เผยแพร่เมื่อ 14 ต.ค. 2024
- Meet In The Middle Technique | Leetcode Partition Array Into Two Arrays to Minimize Sum Difference | CSES Problem set
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was initially trying to solve this problem using memoization but it was showing memory limit exceeded but this approach was better, thanks anmol bhaiya :)
Thanks man
Love you, thank you for the detail explanation.
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;
did you solve the whole cses problem sheet?
Tried to solve the maximum I could
This is really good and thorough! Kudos Anmol!! Keep Going!
Thanks man
Which software are you using to write?
Obs
Good notes
nice video very helpful
in the formula sum-2*(sum1) , why u wrote 1+sum2
same doubt , in example it should be 22 - 2*(3 + 7) = 2
Bhai kya he sahi socha h Awesome !! kitna practice kiya ho bhai itna soch lane ke liye ?
Bhai tejj hi hoo bachpan sa mein😂😂 jokes apart I have been doing dsa for past 4 years.
nice video understood nicely :)
Thanks man
can you explain sum - 2*(1 + sum2)
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 < (1 0){
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 < (1 0){
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
full confusion bro
Watch it again
elements can be negative also, so ig your solution is wrong.
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
@@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?
@@adityaagarwal1650 even I did added that condition previously but test cases were like that my solution passed