// 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"); }