Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in your array.
For example, the length of your array of zeros . Your list of queries is as follows:
a b k 1 5 3 4 8 7 6 9 1
Add the values of between the indices and inclusive:
index-> 1 2 3 4 5 6 7 8 9 10 [0,0,0, 0, 0,0,0,0,0, 0] [3,3,3, 3, 3,0,0,0,0, 0] [3,3,3,10,10,7,7,7,0, 0] [3,3,3,10,10,8,8,8,1, 0]
The largest value is after all operations are performed.
Complete the function arrayManipulation in the editor below. It must return an integer, the maximum value in the resulting array.
arrayManipulation has the following parameters:
- n - the number of elements in your array
- queries - a two dimensional array of queries where each queries[i] contains three integers, a, b, and k.
The first line contains two space-separated integers and , the size of the array and the number of operations.
Each of the next lines contains three space-separated integers , and , the left index, right index and summand.
Return the integer maximum value in the finished array.
5 3 1 2 100 2 5 100 3 4 100
After the first update list will be
100 100 0 0 0.
After the second update list will be
100 200 100 100 100.
After the third update list will be
100 200 200 200 100.
The required answer will be .
Solution in C++
Solution in Java
Solution in Python
from collections import Counter def arrayManipulation(n, queries): c = Counter() for a,b,k in queries: c[a] +=k c[b+1]-=k arrSum = 0 maxSum = 0 for i in sorted(c)[:-1]: arrSum+= c[i] maxSum = max(maxSum,arrSum) return maxSum n,m = map(int,input().split()) queries = [list(map(int,input().split())) for _ in range(m)] print(arrayManipulation(n, queries))
- Given a range[a, b] and a value k we need to add k to all the numbers whose indices are in the range from [a, b].
- We can do an O(1) update by adding to index a and add -k to index b+1.
- Doing this kind of update, the ith number in the array will be prefix sum of array from index 1 to i because we are adding k to the value at index b+1 and subtracting from the value at index and taking prefix sum will give us the actual value for each index after m operations .
- So, we can do all m updates in O(m) time. Now we have to check the largest number in the original array. i.e. the index i such that prefix sum attains the maximum value.
- We can calculate all prefix sums as well as maximum prefix sum in O(n) time which will execute in time.