# Editing Prefix sum array and difference array

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To solve this, we will first transform array <math>S</math> into its prefix sum array <math>P(0,S)</math>. Notice that the sum of each contiguous subsequence <math>S_i + S_{i+1} + S_{i+2} + ... + S_{j-1}</math> corresponds to the difference of two elements of <math>P</math>, that is, <math>P_j - P_i</math>. So what we want to find is the number of pairs <math>(i,j)</math> with <math>P_j - P_i = 47</math> and <math>i < j</math>. (Note that if <math>i > j</math>, we will instead get a subsequence with sum -47.) | To solve this, we will first transform array <math>S</math> into its prefix sum array <math>P(0,S)</math>. Notice that the sum of each contiguous subsequence <math>S_i + S_{i+1} + S_{i+2} + ... + S_{j-1}</math> corresponds to the difference of two elements of <math>P</math>, that is, <math>P_j - P_i</math>. So what we want to find is the number of pairs <math>(i,j)</math> with <math>P_j - P_i = 47</math> and <math>i < j</math>. (Note that if <math>i > j</math>, we will instead get a subsequence with sum -47.) | ||

− | However, this is quite easy to do. We sweep through <math>P</math> from left to right, keeping a [[map]] of all elements of <math>P</math> we've seen so far, along with their frequencies; and for each element <math>P_j</math> we count the number of times <math>P_j - | + | However, this is quite easy to do. We sweep through <math>P</math> from left to right, keeping a [[map]] of all elements of <math>P</math> we've seen so far, along with their frequencies; and for each element <math>P_j</math> we count the number of times <math>P_j - 48</math> has appeared so far, by looking up that value in our map; this tells us how many contiguous subsequences ending at <math>S_{j-1}</math> have sum 47. And finally, adding the number of contiguous subsequences with sum 47 ending at each entry of <math>S</math> gives the total number of such subsequences in the array. Total time taken is <math>O(N)</math>, if we use a [[hash table]] implementation of the map. |

==Use of difference array== | ==Use of difference array== | ||

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After the prefix sum array has been computed, we can use it to add together any rectangle in <math>A</math>, that is, all the elements with their indices in <math>[x_1,x_2), [y_1,y_2)</math>. To do so, we first observe that <math>P_{x_2,y_2} - P_{x_1,y_2}</math> gives the sum of all the elements with indices in the box <math>[x_1, x_2) \times [0, y_2)</math>. Then we similarly observe that <math>P_{x_2,y_1} - P_{x_1,y_1}</math> corresponds to the box <math>[x_1, x_2) \times [0, y_1)</math>. Subtracting gives the box <math>[x_1, x_2) \times [y_1, y_2)</math>. This gives a final formula of <math>P_{x_2,y_2} - P_{x_1,y_2} - P_{x_2,y_1} + P_{x_1,y_1}</math>. | After the prefix sum array has been computed, we can use it to add together any rectangle in <math>A</math>, that is, all the elements with their indices in <math>[x_1,x_2), [y_1,y_2)</math>. To do so, we first observe that <math>P_{x_2,y_2} - P_{x_1,y_2}</math> gives the sum of all the elements with indices in the box <math>[x_1, x_2) \times [0, y_2)</math>. Then we similarly observe that <math>P_{x_2,y_1} - P_{x_1,y_1}</math> corresponds to the box <math>[x_1, x_2) \times [0, y_1)</math>. Subtracting gives the box <math>[x_1, x_2) \times [y_1, y_2)</math>. This gives a final formula of <math>P_{x_2,y_2} - P_{x_1,y_2} - P_{x_2,y_1} + P_{x_1,y_1}</math>. | ||

− | In the <math>n</math>-dimensional case, to sum the elements of <math>A</math> with indices in the box <math>[x_{1,0}, x_{1,1}) \times [x_{2,0}, x_{2,1}) \times ... \times [x_{n,0}, x_{n,1})</math>, we will use the formula <math>\sum_{k_1=0}^1 \sum_{k_2=0}^1 ... \sum_{k_n=0}^1 | + | In the <math>n</math>-dimensional case, to sum the elements of <math>A</math> with indices in the box <math>[x_{1,0}, x_{1,1}) \times [x_{2,0}, x_{2,1}) \times ... \times [x_{n,0}, x_{n,1})</math>, we will use the formula <math>(-1)^n \sum_{k_1=0}^1 \sum_{k_2=0}^1 ... \sum_{k_n=0}^1 P_{x_{1,k_1}, x_{2,k_2}, ..., x_{n,k_n}}</math>. We will not state the proof, instead noting that it is a form of the inclusion–exclusion principle. |

====Example: Diamonds (BOI)==== | ====Example: Diamonds (BOI)==== | ||

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We will simply define the difference array <math>D</math> to be the array whose prefix sum array is <math>A</math>. This means in particular that <math>D_{i,j}</math> is the sum of elements of <math>D</math> with indices in the box <math>[i,i+1) \times [j,j+1)</math> (this box contains only the single element <math>D_{i,j}</math>). Using the prefix sum array <math>A</math>, we obtain <math>D_{i,j} = A_{i+1,j+1} - A_{i,j+1} - A_{i+1,j} + A_{i,j}</math>. In general: | We will simply define the difference array <math>D</math> to be the array whose prefix sum array is <math>A</math>. This means in particular that <math>D_{i,j}</math> is the sum of elements of <math>D</math> with indices in the box <math>[i,i+1) \times [j,j+1)</math> (this box contains only the single element <math>D_{i,j}</math>). Using the prefix sum array <math>A</math>, we obtain <math>D_{i,j} = A_{i+1,j+1} - A_{i,j+1} - A_{i+1,j} + A_{i,j}</math>. In general: | ||

− | :<math>D_{i_1, i_2, ..., i_n} = \sum_{k_1=0}^1 \sum_{k_2=0}^1 ... \sum_{k_n=0}^1 | + | :<math>D_{i_1, i_2, ..., i_n} = (-1)^n \sum_{k_1=0}^1 \sum_{k_2=0}^1 ... \sum_{k_n=0}^1 A_{i_1+k_1,i_2+k_2,...,i_n+k_n}</math>. |

Should we actually need to ''compute'' the difference array, on the other hand, the easiest way to do so is by reversing the computation of the prefix sum array: | Should we actually need to ''compute'' the difference array, on the other hand, the easiest way to do so is by reversing the computation of the prefix sum array: |