Luck Balance | Solution using C | HackerRank

Statement

Lena is preparing for an important coding competition that is preceded by a number of sequential preliminary contests. Initially, her luck balance is 0. She believes in "saving luck", and wants to check her theory. Each contest is described by two integers,  L[i] and T[i]:

• L[i] is the amount of luck associated with a contest. If Lena wins the contest, her luck balance will decrease by L[i]; if she loses it, her luck balance will increase by L[i].
• T[i] denotes the contest's importance rating. It's equal to 1 if the contest is important, and it's equal to 0 if it's unimportant.

If Lena loses no more than  important contests, what is the maximum amount of luck she can have after competing in all the preliminary contests? This value may be negative.

Example

K = 2

L = [5,1,4]

T = [1,2,0]

Contest	L[i]	T[i]
1	5	1
2	1	1
3	4	0

If Lena loses all of the contests, her will be 5+4+1 = 10. Since she is allowed to lose 2 important contests, and there are only 2 important contests, she can lose all three contests to maximize her luck at 10.

If k=1, she has to win at least 1 of the 2 important contests. She would choose to win the lowest value important contest worth 1. Her final luck will be 5+4-1=8.

Function Description

Complete the luckBalance function in the editor below.

luckBalance has the following parameter(s):

• int k: the number of important contests Lena can lose
• int contests[n][2]: a 2D array of integers where each  contains two integers that represent the luck balance and importance of the ith contest

Returns

• int: the maximum luck balance achievable

Solution in C:

/*
 * Complete the 'luckBalance' function below.
 *
 * The function is expected to return an INTEGER.
 * The function accepts following parameters:
 *  1. INTEGER k
 *  2. 2D_INTEGER_ARRAY contests
 */
 
int compare(const void *a, const void *b){
    return (*(int*)b-*(int*)a);
}

int luckBalance(int k, int contests_rows, int contests_columns, int** contests) {
    
    int imp_contest[100];
    int imp_contest_length = 0;
    int luckBalance = 0;
    for (int i=0; i<contests_rows; i++) {
        if(contests[i][1]==1){
            
            imp_contest[imp_contest_length]=contests[i][0];
            imp_contest_length++;
        }
        else {
            luckBalance+=contests[i][0];
        }
        
    }

    qsort(imp_contest, imp_contest_length, sizeof(int), compare);
    printf("%d %d\n",k,imp_contest_length);
    int n = (k<imp_contest_length)?k:imp_contest_length;
    printf("%d",n);
    for (int  i=0; i<n; i++) {
        luckBalance += imp_contest[i];
    }
    for (int i=k; i<imp_contest_length; i++) {
        luckBalance-=imp_contest[i];
    }
    
    return luckBalance;
}

//Solution End here. Driver code starts below ;)

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