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

**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++

`// C++ implementation of the above approach ` ` ` `#include <bits/stdc++.h> ` `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; ` `} ` |

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## Java

`// 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.. ` |

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## Python3

`# 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 ` |

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## C#

`// 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.. ` |

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**Output:**

52

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