github Portfolio Balances An investor opens a new account and wants to invest in a number of assets. Each asset begins with a ba

Question

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github Portfolio Balances An investor opens a new account and wants to invest in a number of assets. Each asset begins with a ba

lance of o, and its value is stored in an array using 7-based indexing. Periodically, a contribution is received and equal investments are made in a subset of the portfolio. Each contribution will be given by investment amount, start index, end index. Each investment in that range will receive the contribution amount. Determine the maximum amount invested in any one investment after all contributions

Computers and Technology

1 answer:

Rzqust [894] · Answer 1 · 2026-02-18T09:47:21+03:00

This code serves to find out the highest amount invested

Explanation:

long maxValue(int n, int rounds_rows, int rounds_columns, int into rounds)

{

// Variable declaration to hold

// the highest investment amount.

long max = 0;

// Creating an array of size n,

// to keep track of the n investments.

long *investments = (long*)malloc(sizeof(long)*n);

int i=0;

// Initialize all

// investments to zero.

for(i=0;i<n;i++)

{

investments[i] = 0;

}

i=0;

// Execute the loop to

// conduct the rounds.

while(i<rounds_rows)

{

// Acquire the left value

// for the current round.

int left = rounds[i][0];

// Acquire the right value

// for this round.

int right = rounds[i][1];

// Get the contribution

// for this round.

int contribution = rounds[i][2];

// Because user indexing is 1-based,

// subtract 1 from left

// and right as the program employs

// 0-based indexing. The

// array investments begins

// at 0 and not 1.

right = right - 1;

int j=0;

// Execute loop to distribute the

// contribution across all investments

// from left to right, inclusive.

for(j=left; j<=right; j++)

{

investments[j] += contribution;

}

i++;

}

// Traverse the investments array

// to locate the maximum value.

max = investments[0];

for(i=1; i<n;i++)

{

if(investments[i]>max)

{

max = investments[i];

}

// Return the

// highest investment.

return max;

}