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authorJack Sangdahl <[email protected]>2025-03-26 20:36:50 -0600
committerJack Sangdahl <[email protected]>2025-03-26 20:36:50 -0600
commit3b7e5a469d82048dbb0122dd2207390ef8f5f653 (patch)
tree8de171df3d9ddab351cae84a6ce0e46346bd5f72 /min_idx_valid_split.c
parent4772cdabe0a208eca2e63d5eb56b3f0bc2164eeb (diff)
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2780. minimum index of a valid split
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+// 2780. Minimum Index of a Valid Split
+#include "leetcode.h"
+
+/*
+ * an element 'x' of an integer array 'arr' of length 'm' is dominant if more
+ * than half the elements of 'arr' have a value of 'x'. you are given a
+ * 0-indexed integer array 'nums' of length 'n' with one dominant element. you
+ * can split 'nums' at an index 'i' into two arrays 'nums[0, ..., i]' and
+ * 'nums[i + 1, ..., n - 1]', but the split is only valid if '0 <= i < - 1' and
+ * the two arrays have the same dominant element. return the minimum index of a
+ * valid split. if no split exists, return -1.
+ */
+
+int minimumIndex(int *nums, int numsSize) {
+ int dominant = nums[0], cnt = 0;
+ for (int i = 0; i < numsSize; i++) {
+ cnt = nums[i] == dominant ? cnt + 1 : cnt - 1;
+ if (!cnt) {
+ dominant = nums[i];
+ cnt = 1;
+ }
+ }
+ int left = 0;
+ for (int i = 0; i < numsSize; i++)
+ if (nums[i] == dominant)
+ left++;
+ cnt = 0;
+ for (int i = 0; i < numsSize; i++) {
+ if (nums[i] == dominant)
+ cnt++;
+ if (cnt * 2 > (i + 1) && (left - cnt) * 2 > (numsSize - i - 1))
+ return i;
+ }
+ return -1;
+}
+
+int main() {
+ int n1[] = {1, 2, 2, 2}, n2[] = {2, 1, 3, 1, 1, 1, 7, 1, 2, 1},
+ n3[] = {3, 3, 3, 3, 7, 2, 2};
+ printf("%d\n", minimumIndex(n1, ARRAY_SIZE(n1))); // expect: 2
+ printf("%d\n", minimumIndex(n2, ARRAY_SIZE(n2))); // expect: 4
+ printf("%d\n", minimumIndex(n3, ARRAY_SIZE(n3))); // expect: -1
+}