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-rw-r--r--minimum_array_end.c26
-rw-r--r--minimum_array_end.py32
2 files changed, 58 insertions, 0 deletions
diff --git a/minimum_array_end.c b/minimum_array_end.c
new file mode 100644
index 0000000..f71563b
--- /dev/null
+++ b/minimum_array_end.c
@@ -0,0 +1,26 @@
+// 3133. Minimum Array End
+#include "leetcode.h"
+
+/*
+ * you are given two integers 'n' and 'x'. you have to construct an array of
+ * positive integers 'nums'. of size 'n' where for every '0 <= i < n - 1',
+ * 'nums[i + 1]' is greater than 'nums[i]', and the result of the bitwise and
+ * operation between all elements of 'nums' is 'x'. return the minimum possible
+ * value of 'nums[n - 1]'
+ */
+
+long long minEnd(int n, int x) {
+ n--;
+ long long a = x, b;
+ for (b = 1; n; b <<= 1)
+ if (!(b & x)) {
+ a |= (n & 1) * b;
+ n >>= 1;
+ }
+ return a;
+}
+
+int main() {
+ printf("%lld\n", minEnd(3, 4)); // expect: 6
+ printf("%lld\n", minEnd(2, 7)); // expect: 15
+}
diff --git a/minimum_array_end.py b/minimum_array_end.py
new file mode 100644
index 0000000..67240ae
--- /dev/null
+++ b/minimum_array_end.py
@@ -0,0 +1,32 @@
+# 3133. Minimum Array End
+
+"""
+you are given two integers 'n' and 'x'. you have to construct an array of
+positive integers 'nums'. of size 'n' where for every '0 <= i < n - 1',
+'nums[i + 1]' is greater than 'nums[i]', and the result of the bitwise and
+operation between all elements of 'nums' is 'x'. return the minimum possible
+value of 'nums[n - 1]'
+"""
+
+
+class Solution(object):
+ def minEnd(self, n, x):
+ """
+ :type n: int
+ :type x: int
+ :rtype: int
+ """
+ n -= 1
+ b = 1
+ for i in range(64):
+ if b & x == 0:
+ x |= (n & 1) * b
+ n >>= 1
+ b <<= 1
+ return x
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.minEnd(3, 4))
+ print(obj.minEnd(2, 7))