Given a matrix (or 2D array) a of integers, find prefix sum matrix for it. Let prefix sum matrix be psa. The value of psa[i][j] contains sum of all values which are above it or on left of it.
Prerequisite: Prefix Sum – 1D
A simple solution is to find psa[i][j] by traversing and adding values from a to a[i][j]. Time complexity o this solution is O(R * C * R * C).
An efficient solution is to use previously computed values to compute psa[i][j]. Unlike 1D array prefix sum, this is tricky, here if we simply add psa[i][j-1] and psa[i-1][j], we get sum of elements from a to a[i-1][j-1] twice, so we subtract psa[i-1][j-1].
psa = psa + psa - psa + a = 6 + 6 - 4 + 1 = 9 The general formula: psa[i][j] = psa[i-1][j] + psa[i][j-1] - psa[i-1][j-1] + a[i][j] Corner Cases (First row and first column) If i = 0 and j = 0 psa[i][j] = a[i][j] If i = 0 and j > 0 psa[i][j] = psa[i][j-1] + a[i][j] If i > 0 and j = 0 psa[i][j] = psa[i-1][j] + a[i][j]
Below is the implementation of the above approach
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Time Complexity : O(R*C)
Auxiliary Space : O(R*C)
- Rearrange the array to maximize the number of primes in prefix sum of the array
- Count the number of primes in the prefix sum array of the given array
- Maximize the sum of array by multiplying prefix of array with -1
- Maximum sum increasing subsequence from a prefix and a given element after prefix is must
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- Prefix Sum Array - Implementation and Applications in Competitive Programming
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- Maximum prefix-sum for a given range
- Postfix to Prefix Conversion
- Check if matrix can be converted to another matrix by transposing square sub-matrices
- Longest Common Prefix Matching | Set-6
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Improved By : manishshaw1