# Minimize the sum after choosing elements from the given three arrays

Given three arrays A[], B[] and C[] of same size N. The task is to minimize the sum after choosing N elements from these array such that at every index i an element from any one of the array A[i], B[i] or C[i] can be chosen and no two consecutive elements can be chosen from the same array.

Examples:

Input: A[] = {1, 100, 1}, B[] = {100, 100, 100}, C[] = {100, 100, 100}
Output: 102
A[0] + B[1] + A[2] = 1 + 100 + 100 = 201
A[0] + B[1] + C[2] = 1 + 100 + 100 = 201
A[0] + C[1] + B[2] = 1 + 100 + 100 = 201
A[0] + C[1] + A[2] = 1 + 100 + 1 = 102
B[0] + A[1] + B[2] = 100 + 100 + 100 = 300
B[0] + A[1] + C[2] = 100 + 100 + 100 = 300
B[0] + C[1] + A[2] = 100 + 100 + 1 = 201
B[0] + C[1] + B[2] = 100 + 100 + 100 = 300
C[0] + A[1] + B[2] = 100 + 100 + 100 = 300
C[0] + A[1] + C[2] = 100 + 100 + 100 = 300
C[0] + B[1] + A[2] = 100 + 100 + 1 = 201
C[0] + B[1] + C[2] = 100 + 100 + 100 = 300

Input: A[] = {1, 1, 1}, B[] = {1, 1, 1}, C[] = {1, 1, 1}
Output: 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The problem is a simple variation of finding minimum cost. The extra constraint are that if we take an element from a particular array then we cannot take the next element from the same array. This could easily be solved using recursion but it would give time complexity as O(3^n) because for every element we have three arrays as choices.

To improve the time complexity we can easily store the pre-calculated values in a dp array.

Since there are three arrays to choose from at every index, three cases arise in this scenario:

• Case 1: If array A[] is selected from the ith element then we either choose the array B[] or the array C[] for the (i + 1)th element.
• Case 2: If array B[] is selected from the ith element then we either choose the array A[] or the array C[] for the (i + 1)th element.
• Case 3: If array C[] is selected from the ith element then we either choose the array A[] or the array B[] for the (i + 1)th element.

The above states can be solved using recursion and intermediate results can be stored in the dp array.

Below is the implementation of the above approach:

 `// C++ implementation of the above approach ` ` `  `#include ` `using` `namespace` `std; ` `#define SIZE 3 ` `const` `int` `N = 3; ` ` `  `// Function to return the minimized sum ` `int` `minSum(``int` `A[], ``int` `B[], ``int` `C[], ``int` `i, ` `        ``int` `n, ``int` `curr, ``int` `dp[SIZE][N]) ` `{ ` ` `  `    ``// If all the indices have been used ` `    ``if` `(n <= 0) ` `        ``return` `0; ` ` `  `    ``// If this value is pre-calculated ` `    ``// then return its value from dp array ` `    ``// instead of re-computing it ` `    ``if` `(dp[n][curr] != -1) ` `        ``return` `dp[n][curr]; ` ` `  `    ``// Here curr is the array chosen ` `    ``// for the (i - 1)th element ` `    ``// 0 for A[], 1 for B[] and 2 for C[] ` ` `  `    ``// If A[i - 1] was chosen previously then ` `    ``// only B[i] or C[i] can chosen now ` `    ``// choose the one which leads ` `    ``// to the minimum sum ` `    ``if` `(curr == 0) { ` `        ``return` `dp[n][curr]  ` `                ``= min(B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp), ` `                      ``C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp)); ` `    ``} ` ` `  `    ``// If B[i - 1] was chosen previously then ` `    ``// only A[i] or C[i] can chosen now ` `    ``// choose the one which leads ` `    ``// to the minimum sum ` `    ``if` `(curr == 1) ` `        ``return` `dp[n][curr]  ` `                ``= min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp), ` `                      ``C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp)); ` ` `  `    ``// If C[i - 1] was chosen previously then ` `    ``// only A[i] or B[i] can chosen now ` `    ``// choose the one which leads ` `    ``// to the minimum sum ` `    ``return` `dp[n][curr]  ` `                ``= min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp), ` `                      ``B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `A[] = { 1, 50, 1 }; ` `    ``int` `B[] = { 50, 50, 50 }; ` `    ``int` `C[] = { 50, 50, 50 }; ` ` `  `    ``// Initialize the dp[][] array ` `    ``int` `dp[SIZE][N]; ` `    ``for` `(``int` `i = 0; i < SIZE; i++) ` `        ``for` `(``int` `j = 0; j < N; j++) ` `            ``dp[i][j] = -1; ` ` `  `    ``// min(start with A[0], start with B[0], start with C[0]) ` `    ``cout << min(A[0] + minSum(A, B, C, 1, SIZE - 1, 0, dp), ` `                ``min(B[0] + minSum(A, B, C, 1, SIZE - 1, 1, dp), ` `                    ``C[0] + minSum(A, B, C, 1, SIZE - 1, 2, dp))); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the above approach ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` ` `  `static` `int` `SIZE = ``3``; ` `static` `int` `N = ``3``; ` ` `  `// Function to return the minimized sum ` `static` `int` `minSum(``int` `A[], ``int` `B[], ``int` `C[], ``int` `i, ` `                    ``int` `n, ``int` `curr, ``int` `[][]dp) ` `{ ` ` `  `    ``// If all the indices have been used ` `    ``if` `(n <= ``0``) ` `        ``return` `0``; ` ` `  `    ``// If this value is pre-calculated ` `    ``// then return its value from dp array ` `    ``// instead of re-computing it ` `    ``if` `(dp[n][curr] != -``1``) ` `        ``return` `dp[n][curr]; ` ` `  `    ``// Here curr is the array chosen ` `    ``// for the (i - 1)th element ` `    ``// 0 for A[], 1 for B[] and 2 for C[] ` ` `  `    ``// If A[i - 1] was chosen previously then ` `    ``// only B[i] or C[i] can chosen now ` `    ``// choose the one which leads ` `    ``// to the minimum sum ` `    ``if` `(curr == ``0``)  ` `    ``{ ` `        ``return` `dp[n][curr]  ` `                ``= Math.min(B[i] + minSum(A, B, C, i + ``1``, n - ``1``, ``1``, dp), ` `                    ``C[i] + minSum(A, B, C, i + ``1``, n - ``1``, ``2``, dp)); ` `    ``} ` ` `  `    ``// If B[i - 1] was chosen previously then ` `    ``// only A[i] or C[i] can chosen now ` `    ``// choose the one which leads ` `    ``// to the minimum sum ` `    ``if` `(curr == ``1``) ` `        ``return` `dp[n][curr]  ` `                ``= Math.min(A[i] + minSum(A, B, C, i + ``1``, n - ``1``, ``0``, dp), ` `                    ``C[i] + minSum(A, B, C, i + ``1``, n - ``1``, ``2``, dp)); ` ` `  `    ``// If C[i - 1] was chosen previously then ` `    ``// only A[i] or B[i] can chosen now ` `    ``// choose the one which leads ` `    ``// to the minimum sum ` `    ``return` `dp[n][curr]  ` `                ``= Math.min(A[i] + minSum(A, B, C, i + ``1``, n - ``1``, ``0``, dp), ` `                    ``B[i] + minSum(A, B, C, i + ``1``, n - ``1``, ``1``, dp)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `A[] = { ``1``, ``50``, ``1` `}; ` `    ``int` `B[] = { ``50``, ``50``, ``50` `}; ` `    ``int` `C[] = { ``50``, ``50``, ``50` `}; ` `     `  `    ``// Initialize the dp[][] array ` `    ``int` `dp[][] = ``new` `int``[SIZE][N]; ` `    ``for` `(``int` `i = ``0``; i < SIZE; i++) ` `        ``for` `(``int` `j = ``0``; j < N; j++) ` `            ``dp[i][j] = -``1``; ` `     `  `    ``// min(start with A[0], start with B[0], start with C[0]) ` `    ``System.out.println(Math.min(A[``0``] + minSum(A, B, C, ``1``, SIZE - ``1``, ``0``, dp), ` `                ``Math.min(B[``0``] + minSum(A, B, C, ``1``, SIZE - ``1``, ``1``, dp), ` `                    ``C[``0``] + minSum(A, B, C, ``1``, SIZE - ``1``, ``2``, dp)))); ` `} ` `} ` ` `  `// This code is contributed by anuj_67..  `

 `# Python3 implementation of the above approach  ` ` `  `import` `numpy as np ` ` `  `SIZE ``=` `3``;  ` `N ``=` `3``;  ` ` `  `# Function to return the minimized sum  ` `def` `minSum(A, B, C, i, n, curr, dp) :  ` ` `  `    ``# If all the indices have been used  ` `    ``if` `(n <``=` `0``) : ` `        ``return` `0``;  ` ` `  `    ``# If this value is pre-calculated  ` `    ``# then return its value from dp array  ` `    ``# instead of re-computing it  ` `    ``if` `(dp[n][curr] !``=` `-``1``) : ` `        ``return` `dp[n][curr];  ` ` `  `    ``# Here curr is the array chosen  ` `    ``# for the (i - 1)th element  ` `    ``# 0 for A[], 1 for B[] and 2 for C[]  ` ` `  `    ``# If A[i - 1] was chosen previously then  ` `    ``# only B[i] or C[i] can chosen now  ` `    ``# choose the one which leads  ` `    ``# to the minimum sum  ` `    ``if` `(curr ``=``=` `0``) :  ` `        ``dp[n][curr] ``=` `min``( B[i] ``+` `minSum(A, B, C, i ``+` `1``, n ``-` `1``, ``1``, dp),  ` `        ``C[i] ``+` `minSum(A, B, C, i ``+` `1``, n ``-` `1``, ``2``, dp));  ` `        ``return` `dp[n][curr]  ` `     `  ` `  `    ``# If B[i - 1] was chosen previously then  ` `    ``# only A[i] or C[i] can chosen now  ` `    ``# choose the one which leads  ` `    ``# to the minimum sum  ` `    ``if` `(curr ``=``=` `1``) : ` `        ``dp[n][curr] ``=` `min``(A[i] ``+` `minSum(A, B, C, i ``+` `1``, n ``-` `1``, ``0``, dp),  ` `        ``C[i] ``+` `minSum(A, B, C, i ``+` `1``, n ``-` `1``, ``2``, dp));  ` `        ``return` `dp[n][curr] ` ` `  `    ``# If C[i - 1] was chosen previously then  ` `    ``# only A[i] or B[i] can chosen now  ` `    ``# choose the one which leads  ` `    ``# to the minimum sum  ` `    ``dp[n][curr] ``=` `min``(A[i] ``+` `minSum(A, B, C, i ``+` `1``, n ``-` `1``, ``0``, dp),  ` `    ``B[i] ``+` `minSum(A, B, C, i ``+` `1``, n ``-` `1``, ``1``, dp));  ` `     `  `    ``return` `dp[n][curr] ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``A ``=` `[ ``1``, ``50``, ``1` `];  ` `    ``B ``=` `[ ``50``, ``50``, ``50` `];  ` `    ``C ``=` `[ ``50``, ``50``, ``50` `];  ` ` `  `    ``# Initialize the dp[][] array  ` `    ``dp ``=` `np.zeros((SIZE,N));  ` `     `  `    ``for` `i ``in` `range``(SIZE) : ` `        ``for` `j ``in` `range``(N) :  ` `            ``dp[i][j] ``=` `-``1``;  ` ` `  `    ``# min(start with A[0], start with B[0], start with C[0])  ` `    ``print``(``min``(A[``0``] ``+` `minSum(A, B, C, ``1``, SIZE ``-` `1``, ``0``, dp),  ` `                ``min``(B[``0``] ``+` `minSum(A, B, C, ``1``, SIZE ``-` `1``, ``1``, dp),  ` `                ``C[``0``] ``+` `minSum(A, B, C, ``1``, SIZE ``-` `1``, ``2``, dp))));  ` ` `  `# This code is contributed by AnkitRai01 `

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `static` `int` `SIZE = 3; ` `static` `int` `N = 3; ` ` `  `// Function to return the minimized sum ` `static` `int` `minSum(``int` `[]A, ``int` `[]B, ``int` `[]C, ``int` `i, ` `                    ``int` `n, ``int` `curr, ``int` `[,]dp) ` `{ ` ` `  `    ``// If all the indices have been used ` `    ``if` `(n <= 0) ` `        ``return` `0; ` ` `  `    ``// If this value is pre-calculated ` `    ``// then return its value from dp array ` `    ``// instead of re-computing it ` `    ``if` `(dp[n,curr] != -1) ` `        ``return` `dp[n,curr]; ` ` `  `    ``// Here curr is the array chosen ` `    ``// for the (i - 1)th element ` `    ``// 0 for A[], 1 for B[] and 2 for C[] ` ` `  `    ``// If A[i - 1] was chosen previously then ` `    ``// only B[i] or C[i] can chosen now ` `    ``// choose the one which leads ` `    ``// to the minimum sum ` `    ``if` `(curr == 0)  ` `    ``{ ` `        ``return` `dp[n,curr]  ` `                ``= Math.Min(B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp), ` `                    ``C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp)); ` `    ``} ` ` `  `    ``// If B[i - 1] was chosen previously then ` `    ``// only A[i] or C[i] can chosen now ` `    ``// choose the one which leads ` `    ``// to the minimum sum ` `    ``if` `(curr == 1) ` `        ``return` `dp[n,curr]  ` `                ``= Math.Min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp), ` `                    ``C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp)); ` ` `  `    ``// If C[i - 1] was chosen previously then ` `    ``// only A[i] or B[i] can chosen now ` `    ``// choose the one which leads ` `    ``// to the minimum sum ` `    ``return` `dp[n,curr]  ` `                ``= Math.Min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp), ` `                    ``B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `[]A = { 1, 50, 1 }; ` `    ``int` `[]B = { 50, 50, 50 }; ` `    ``int` `[]C = { 50, 50, 50 }; ` `     `  `    ``// Initialize the dp[][] array ` `    ``int` `[,]dp = ``new` `int``[SIZE,N]; ` `    ``for` `(``int` `i = 0; i < SIZE; i++) ` `        ``for` `(``int` `j = 0; j < N; j++) ` `            ``dp[i,j] = -1; ` `     `  `    ``// min(start with A[0], start with B[0], start with C[0]) ` `    ``Console.WriteLine(Math.Min(A[0] + minSum(A, B, C, 1, SIZE - 1, 0, dp), ` `                ``Math.Min(B[0] + minSum(A, B, C, 1, SIZE - 1, 1, dp), ` `                    ``C[0] + minSum(A, B, C, 1, SIZE - 1, 2, dp)))); ` `} ` `} ` ` `  `// This code is contributed by anuj_67..  `

Output:
```52
```

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