Data Modelling & AI Data Structure & Algorithm

Minimize the sum after choosing elements from the given three arrays

23 June 2025

0

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;
}

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

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

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

Create a DP to store the solution of the subproblems.
Initialize the DP with base cases
Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP.
Return the final solution minimum of min(dp[0][0], min(dp[0][1], dp[0][2])).

Implementation :

C++

#include <iostream>
using namespace std;
 
const int SIZE = 3;
 
// Function to return the minimized sum
int minSum(int A[], int B[], int C[]) {
    int dp[SIZE][3];
 
    // Base case
    dp[SIZE-1][0] = A[SIZE-1];
    dp[SIZE-1][1] = B[SIZE-1];
    dp[SIZE-1][2] = C[SIZE-1];
 
    // Tabulate the solution from bottom up
    for(int i = SIZE-2; i >= 0; i--) {
       
          // iterate over subproblems to get the current
          // value from previous computaions
        dp[i][0] = A[i] + min(dp[i+1][1], dp[i+1][2]);
        dp[i][1] = B[i] + min(dp[i+1][0], dp[i+1][2]);
        dp[i][2] = C[i] + min(dp[i+1][0], dp[i+1][1]);
    }
 
    // Return the minimized sum
    return min(dp[0][0], min(dp[0][1], dp[0][2]));
}
 
// Driver code
int main() {
    int A[] = { 1, 50, 1 };
    int B[] = { 50, 50, 50 };
    int C[] = { 50, 50, 50 };
     
      // function call
    cout << minSum(A, B, C) << endl;
    return 0;
}

Java

// Java program to return the minimized sum
public class Main {
    public static void main(String[] args) {
        int SIZE = 3;
        int[] A = { 1, 50, 1 };
        int[] B = { 50, 50, 50 };
        int[] C = { 50, 50, 50 };
 
        int[][] dp = new int[SIZE][3];
 
        // Base case
        dp[SIZE - 1][0] = A[SIZE - 1];
        dp[SIZE - 1][1] = B[SIZE - 1];
        dp[SIZE - 1][2] = C[SIZE - 1];
 
        // Tabulate the solution from bottom up
        for (int i = SIZE - 2; i >= 0; i--) {
 
            // Iterate over subproblems to get the current
            // value from previous computations
            dp[i][0] = A[i] + Math.min(dp[i + 1][1], dp[i + 1][2]);
            dp[i][1] = B[i] + Math.min(dp[i + 1][0], dp[i + 1][2]);
            dp[i][2] = C[i] + Math.min(dp[i + 1][0], dp[i + 1][1]);
        }
 
        // Return the minimized sum
        int result = Math.min(dp[0][0], Math.min(dp[0][1], dp[0][2]));
        System.out.println(result);
    }
}

Python3

import sys
 
SIZE = 3
 
# Function to return the minimized sum
 
 
def minSum(A, B, C):
    dp = [[0 for j in range(3)] for i in range(SIZE)]
 
    # Base case
    dp[SIZE-1][0] = A[SIZE-1]
    dp[SIZE-1][1] = B[SIZE-1]
    dp[SIZE-1][2] = C[SIZE-1]
 
    # Tabulate the solution from bottom up
    for i in range(SIZE-2, -1, -1):
        # iterate over subproblems to get the current
        # value from previous computaions
        dp[i][0] = A[i] + min(dp[i+1][1], dp[i+1][2])
        dp[i][1] = B[i] + min(dp[i+1][0], dp[i+1][2])
        dp[i][2] = C[i] + min(dp[i+1][0], dp[i+1][1])
 
    # Return the minimized sum
    return min(dp[0][0], min(dp[0][1], dp[0][2]))
 
 
# Driver code
if __name__ == '__main__':
    A = [1, 50, 1]
    B = [50, 50, 50]
    C = [50, 50, 50]
 
    # function call
    print(minSum(A, B, C))

C#

using System;
 
class Program
{
    const int SIZE = 3;
 
    // Function to return the minimized sum
    static int MinSum(int[] A, int[] B, int[] C)
    {
        int[,] dp = new int[SIZE, 3];
 
        // Base case
        dp[SIZE - 1, 0] = A[SIZE - 1];
        dp[SIZE - 1, 1] = B[SIZE - 1];
        dp[SIZE - 1, 2] = C[SIZE - 1];
 
        // Tabulate the solution from bottom up
        for (int i = SIZE - 2; i >= 0; i--)
        {
            // Iterate over subproblems to get the current
            // value from previous computations
            dp[i, 0] = A[i] + Math.Min(dp[i + 1, 1], dp[i + 1, 2]);
            dp[i, 1] = B[i] + Math.Min(dp[i + 1, 0], dp[i + 1, 2]);
            dp[i, 2] = C[i] + Math.Min(dp[i + 1, 0], dp[i + 1, 1]);
        }
 
        // Return the minimized sum
        return Math.Min(dp[0, 0], Math.Min(dp[0, 1], dp[0, 2]));
    }
 
    // Driver Code
    static void Main()
    {
        int[] A = { 1, 50, 1 };
        int[] B = { 50, 50, 50 };
        int[] C = { 50, 50, 50 };
 
        // Function call
        Console.WriteLine(MinSum(A, B, C));
    }
}

Javascript

const SIZE = 3;
 
// Function to return the minimized sum
function minSum(A, B, C) {
    const dp = new Array(SIZE).fill().map(() => new Array(3));
 
    // Base case
    dp[SIZE - 1][0] = A[SIZE - 1];
    dp[SIZE - 1][1] = B[SIZE - 1];
    dp[SIZE - 1][2] = C[SIZE - 1];
 
    // Tabulate the solution from bottom up
    for (let i = SIZE - 2; i >= 0; i--) {
        // Iterate over subproblems to get the current
        // value from previous computations
        dp[i][0] = A[i] + Math.min(dp[i + 1][1], dp[i + 1][2]);
        dp[i][1] = B[i] + Math.min(dp[i + 1][0], dp[i + 1][2]);
        dp[i][2] = C[i] + Math.min(dp[i + 1][0], dp[i + 1][1]);
    }
 
    // Return the minimized sum
    return Math.min(dp[0][0], Math.min(dp[0][1], dp[0][2]));
}
 
// Driver code
const A = [1, 50, 1];
const B = [50, 50, 50];
const C = [50, 50, 50];
 
// Function call
console.log(minSum(A, B, C));

Output

Time Complexity: O(n), where n is the size of the arrays A, B, and C.
Auxiliary Space: O(n), as the dp array has n rows and 3 columns

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Minimize the sum after choosing elements from the given three arrays

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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Feedly turned my E Ink tablet into the distraction-free reader I always wanted

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Feedly turned my E Ink tablet into the distraction-free reader I always wanted

OpenAI’s viral video generation app Sora inches closer to an Android release

Pixel 10’s Tensor G5 shows Google still has a lot to learn

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