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// 2028. Find Missing Observations
#include "leetcode.h"
/*
* you have observations of n + m 6 sided dice rolls withh each face numbered
* from 1 to 6. n of the observations went missing and you only have the
* observations of m rolls. fortunately, you have also calculated thhe average
* value of the n + m rolls. you are given an integer array rolls of length m
* where rolls[i] is the value of the i'th observation. you are also given the
* two integers mean and n. return an array of length n containing thhe missing
* observations such that the average value of the n + m rolls is exactly mean.
* if there are multiple valid answers, return any of them. if no such array
* exists, return an empty array. thhe average value of a set of k numbers is
* the sum of the numbers divided by k. note that mean is an integer so the sum
* of the n + m rolls should be divisble by n + m.
*/
int *missingRolls(int *rolls, int rolls_size, int mean, int n,
int *return_size) {
int sum = 0;
for (int i = 0; i < rolls_size; i++)
sum += rolls[i];
int target = mean * (rolls_size + n);
int diff = target - sum;
if (diff < n || diff > 6 * n) {
*return_size = 0;
return NULL;
}
int *ans = malloc(n * sizeof(int));
*return_size = n;
for (int i = 0; i < n; i++)
ans[i] = diff / n;
diff -= n * (diff / n);
int pos = 0;
while (diff) {
if (diff > (6 - ans[pos])) {
diff -= (6 - ans[pos]);
ans[pos] = 6;
} else {
ans[pos] += diff;
diff = 0;
}
pos++;
}
return ans;
}
int main() {
int r1[] = {3, 2, 4, 3}, r2[] = {1, 5, 6}, r3[] = {1, 2, 3, 4};
int rs1[] = {}, rs2[] = {}, rs3[] = {};
int *mr1 = missingRolls(r1, 4, 4, 2, rs1),
*mr2 = missingRolls(r2, 3, 3, 4, rs2),
*mr3 = missingRolls(r3, 4, 6, 4, rs3);
for (int i = 0; i < 2; i++)
printf("%d ", mr1[i]); // expect: 6 6
printf("\n");
for (int i = 0; i < 4; i++)
printf("%d ", mr2[i]); // expect: 2 3 2 2
printf("\n");
for (int i = 0; i < 1; i++)
printf("%d ", mr3[i]); // expect: null
printf("\n");
}
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