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Minimize the sum after choosing elements from the given three arrays

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  • Difficulty Level : Medium
  • Last Updated : 24 Apr, 2022

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:
 

 

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;
}

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..

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

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..

Javascript




<script>
    // Javascript implementation of the above approach
     
    let SIZE = 3;
    let N = 3;
 
    // Function to return the minimized sum
    function 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)
        {
            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));
    }
     
    let A = [ 1, 50, 1 ];
    let B = [ 50, 50, 50 ];
    let C = [ 50, 50, 50 ];
       
    // Initialize the dp[][] array
    let dp = new Array(SIZE);
    for (let i = 0; i < SIZE; i++)
    {
        dp[i] = new Array(N);
        for (let j = 0; j < N; j++)
        {
            dp[i][j] = -1;
        }
    }
       
    // min(start with A[0], start with B[0], start with C[0])
    document.write(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))));
 
</script>

Output: 

52

 

Time Complexity: O(SIZE*N), where dp operations taking SIZE*N time
Auxiliary Space: O(SIZE*N), where dp array is made of two states SIZE*N


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