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authorJack Sangdahl <[email protected]>2025-08-10 11:33:56 +0200
committerJack Sangdahl <[email protected]>2025-08-10 11:33:56 +0200
commit20f07f1feae9cf99ff005d379919fd9bad012f82 (patch)
tree0f7ef975fbf1813fe32b741c161cc08b5473ae10
parenta09548594e7f57c4d6d31964f7c1d99bf2f822b2 (diff)
downloadleetcode-20f07f1feae9cf99ff005d379919fd9bad012f82.tar.gz
leetcode-20f07f1feae9cf99ff005d379919fd9bad012f82.zip
869. reordered power of 2
-rw-r--r--reordered_power_2.c30
-rw-r--r--reordered_power_2.py27
2 files changed, 57 insertions, 0 deletions
diff --git a/reordered_power_2.c b/reordered_power_2.c
new file mode 100644
index 0000000..bdc818a
--- /dev/null
+++ b/reordered_power_2.c
@@ -0,0 +1,30 @@
+// 869. Reordered Power of 2
+#include "leetcode.h"
+
+/*
+ * you are given an integer 'n'. we reorder the digits in any order (including
+ * the original order) such that the leading digit is not zero. return true if
+ * an only if we can do this so that the resulting number is a power of two.
+ */
+
+static uint64_t signature(uint32_t n) {
+ uint64_t res = 0;
+ while (n) {
+ res += 1UL << ((n % 10) * 4);
+ n /= 10;
+ }
+ return res;
+}
+
+bool reorderedPowerOf2(int n) {
+ const uint64_t sig = signature(n);
+ for (int i = 0; i < 32; ++i)
+ if (sig == signature(1 << i))
+ return true;
+ return false;
+}
+
+int main() {
+ printf("%d\n", reorderedPowerOf2(1)); // expect: 1
+ printf("%d\n", reorderedPowerOf2(10)); // expect: 0
+}
diff --git a/reordered_power_2.py b/reordered_power_2.py
new file mode 100644
index 0000000..9318c0c
--- /dev/null
+++ b/reordered_power_2.py
@@ -0,0 +1,27 @@
+# 869. Reordered Power of 2
+from collections import Counter
+
+"""
+you are given an integer 'n'. we reorder the digits in any order (including
+the original order) such that the leading digit is not zero. return true if
+an only if we can do this so that the resulting number is a power of two.
+"""
+
+
+class Solution(object):
+ def reorderedPowerOf2(self, n):
+ """
+ :type n: int
+ :rtype: bool
+ """
+ cnt = Counter(str(n))
+ for i in range(31):
+ if Counter(str(1 << i)) == cnt:
+ return True
+ return False
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.reorderedPowerOf2(1))
+ print(obj.reorderedPowerOf2(10))