revice easy array

revice easy array

September 21, 2023

Q

Approach-

note->

1) we can use the outer loop for take the lenngth of indices
2) we use the [i][0] in first loop to increment the cols by making constant row
3)[i][1] use to change the row by making constant cols

2nd way->

create two boolean array for row and col with row col size
row[indices[i][0]] ^=true; same with col using [1]
then count r and c values ; use following formulae;

Q

sol->

first run loop till length
after that take the sum from primary and secondary side
if length is odd then we have to - the middle element because it comes twice

Q

approch 1->while(0<nums[i]) - count++ nums[i]/=10;
approch2-> Math.log10(nums[i])+1; it produce odd value for even and even for odd so we have to add 1 to it;
approch3-> we set the limit like >9 && < =99 || like wise;

Q

ans->

Use the new 2D ans for [col][row] because if matrix size is 2*4 the after transpose it becomes 4*2 thats why we give the row size depend on cols size.
we use col =matrix[0].length because the size of col in first row is equal to all cols .

Q

solution

In this problem first make a pointer on last index.
add the k value to the last index like {1,2,0,0} + k=34,then 0+34=34%10 add this to list
and make the k value to k=(num[i]+k)/10; to get 34/10= 3 is new k value.
if k>0 means we have {9,9} in this case k=1 remains after all process so we have to add using k%10 and after k/10.
Atlast reverse the list.

Q

In this problem we use a for (i) to go each rows and for(j=logs[i][0] j<logs[i][1]) we have to skip the last year so we go till last year n-1.
after the create one array and increment the index of each index by logs[j-1950]++.
after this one check the max value and after find the first max we return with i+1950 year.
Or we can make at logs[i][0]++ and logs[i][1]-- after that take the running sum apply same process.

Q

we need array to store the two index.
Two ways one we run for loop for i and use another for which is starting from next index.
We can use hashmap to store the key with index as values.

Q

We can approch this problem by two ways 1- by using Math.max directly or 2- using boolean value.
first we have to check at each row so make one row for loop for all operation.
Then first find the min in row and store the col value using that col value we can check at that col making row increment and decrement.

Q

We have to take all the maximum sub array whose sum is greater.
So we have to brute force take the sum of array of first index to till the end and check the max sum between the sum and max if sum<0? sum=0:continue.
Best approach use the loop start from 1-> currentsum=max between(current+current value,current value)
lastsum=max between currentsum, lastsum.

Q

First we check the index is !=9 the simply increment the break, otherwise we have to place the 0 at that index and afterwards make the 0 index value to 1 by increasing the length by 1.

Q

Easy just count the odd and event chips and find the minimum values between them to transfer from one position to another.

Q

First we check the r*c and matrix row and col length if equals the return as it is or size equals the perform action if not then return .
We have two approaches first one we can divide row->[i/col] and for col->[i%col] int this way we find the values.
after atlast return it as it is storing it in new array.

Q

easy to solve

Q

We have to make random array with the sum=0 of length n,So we create the array with length n and just place equal means +1 -1 +2 -2 like wise.

important we increment the array by i+=2 beause we place the values already at the first and second index at one iteration.

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