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This is possible because the algorithm assumes that each of the n input elements is in the range [0, k], i.e. no element is smaller than 0 nor greater than k. The input is an array A indexed from 1 to n, the output is stored in the array B also indexed from 1 to n, and the array C is used as a temporary container, indexed from 0 to k. 1 let C[0..k] be a new array 2 for i = 0 to k 3 C[i] = 0 4 for j = 1 to A.length 5 C[A[j]] = C[A[j]] + 1 6 // C[i] now contains the number of elements equal to i. 7 for i = 1 to k 8 C[i] = C[i] + C[i - 1] 9 // C[i] now contains the number of elements less than or equal to i. 10 for j = A.length downto 1 11 B[C[A[j]]] = A[j] 12 C[A[j]] = C[A[j]] - 1 Given an input array A = [5, 1, 2, 1, 5, 3], what is the status of C at the end of the algorithm? ? 1,2,3,4,4,6 ? none of the others ? 0,1,2,3,5,5 ? 4,4,5,2,0 ? 0,0,2,3,4,4 

          This is possible because the algorithm assumes that each of the n input elements is in the range
[0, k], i.e. no element is smaller than 0 nor greater than k. The input is an array A indexed from 1
to n, the output is stored in the array B also indexed from 1 to n, and the array C is used as a
temporary container, indexed from 0 to k.
1 let C[0..k] be a new array
2 for i = 0 to k
3
C[i] = 0
4 for j = 1 to A.length
5
C[A[j]] = C[A[j]] + 1
6 // C[i] now contains the number of elements equal to i.
7 for i = 1 to k
8
C[i] = C[i] + C[i - 1]
9 // C[i] now contains the number of elements less than or equal to i.
10 for j = A.length downto 1
11
B[C[A[j]]] = A[j]
12
C[A[j]] = C[A[j]] - 1
Given an input array A = [5, 1, 2, 1, 5, 3], what is the status of C at the end of the algorithm?
? 1,2,3,4,4,6
? none of the others
? 0,1,2,3,5,5
? 4,4,5,2,0
? 0,0,2,3,4,4
Show more…
This is possible because the algorithm assumes that each of the n input elements is in the range [0, k], i.e. no element is smaller than 0 nor greater than k. The input is an array A indexed from 1 to n, the output is stored in the array B also indexed from 1 to n, and the array C is used as a temporary container, indexed from 0 to k. 1 let C[0..k] be a new array 2 for i = 0 to k 3 C[i] = 0 4 for j = 1 to A.length 5 C[A[j]] = C[A[j]] + 1 6 // C[i] now contains the number of elements equal to i. 7 for i = 1 to k 8 C[i] = C[i] + C[i - 1] 9 // C[i] now contains the number of elements less than or equal to i. 10 for j = A.length downto 1 11 B[C[A[j]]] = A[j] 12 C[A[j]] = C[A[j]] - 1 Given an input array A = [5, 1, 2, 1, 5, 3], what is the status of C at the end of the algorithm? ? 1,2,3,4,4,6 ? none of the others ? 0,1,2,3,5,5 ? 4,4,5,2,0 ? 0,0,2,3,4,4

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This is possible because the algorithm assumes that each of the n input elements is in the range [0, k], i.e. no element is smaller than 0 nor greater than k. The input is an array A indexed from 1 to n, the output is stored in the array B also indexed from 1 to n, and the array C is used as a temporary container, indexed from 0 to k. Let C[0..k] be a new array. For i = 0 to k: C[i] = 0 For j = 1 to A.length: C[A[j]] = C[A[j]] + 1 // C[i] now contains the number of elements equal to i. For i = 1 to k: C[i] = C[i] + C[i-1] // C[i] now contains the number of elements less than or equal to i. For j = A.length downto 1: B[C[A[j]]] = A[j] C[A[j]] = C[A[j]] - 1
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You are given two arrays A and B consisting of N integers each. An index is named fair if the four sums (A[0] + ... + A[K-1]), (A[K] + ... + A[N-1]), (B[0] + ... + B[K-1]) and (B[K] + ... + B[N-1]) are all equal. In other words, K is the index where the two arrays, A and B, can be split into two non-empty arrays each in such a way that the sums of the resulting arrays' elements are equal. For example, given arrays A = [10, 4, -1, 0, 3] and B = [10, -2, 5, 0, 3], index K=3 is fair. The sums of the subarrays are all equal: 0 + 4 + (-1) = 3, 0 + 3 = 3, 0 + (-2) + 5 = 3, and 0 + 3 = 3. On the other hand, index K=2 is not fair as the sums of the subarrays are: 0 + 4 - 4, (1) + 0 + 3 - 2, 0 + (-2) = 2, and 5 + 0 - 3 - 8. Write a function int solution(int A[], int B[], int N) which, given two arrays of integers A and B, returns the total number of fair indexes. Examples: 1. Given A = [10, 4, -1, 0, 3] and B = [0, -2, 5, 0, 3], your function should return 2. The fair indexes are 3 and 4. In both cases, the sums of elements of the subarrays are equal to 3. 2. Given A = [2, 2, 3, 3] and B = [10, 0, 4, -4], your function should return 1. The only fair index is 2. Index 4 is not fair as the subarrays containing indexes from K to N-1 would be empty. 3. Given A = [14, -10, 3, 1] and B = [12, 6, 0, 4, 1], your function should return 0. There are no fair indexes.

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Transcript

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00:01 Hey, it's clarissa enumerate here.
00:02 So the algorithm only makes comparisons in the line m sub ij is defined to be min of m sub ij, comma, a, sub k.
00:20 So one comparison is made in each iteration of the three four loops.
00:24 So for i, it can take on the values from 1 to n.
00:30 So it is o.
00:32 Hey it's clarissa enumerate here so the algorithm only makes comparisons in the line m sub i j is defined to be min i can take on n values j can take on the values from i plus one to n so j can take on the value take on n minus i values which is at most n minus one values this is assuming the i is equal to n.
01:10 To 1.
01:11 So k can take on the values of 1 plus i to j.
01:15 So j can take on values j minus 1, which is at most n minus 1 values...
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