🚀 06-Jul-2020

Author: sureshmangs

competitive programming/interviewbit/arrays/Spiral Order Matrix I.cpp

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+Given a matrix of m * n elements (m rows, n columns), return all elements of the matrix in spiral order.
+
+Example:
+
+Given the following matrix:
+
+[
+    [ 1, 2, 3 ],
+    [ 4, 5, 6 ],
+    [ 7, 8, 9 ]
+]
+You should return
+
+[1, 2, 3, 6, 9, 8, 7, 4, 5]
+
+
+
+
+
+
+vector<int> Solution::spiralOrder(const vector<vector<int> > &ar) {
+    int m=ar.size();
+    int n=ar[0].size();
+    vector<int>res;
+    int t=0, b=m-1, l=0, r=n-1;
+    int dir=0;  // direction r d l u
+    while(t<=b && l<=r){
+        if(dir==0){
+            for(int i=l; i<=r;i++) res.push_back(ar[t][i]);
+            t++;
+            dir=1;
+            continue;
+        }
+        if(dir==1){
+            for(int i=t; i<=b;i++) res.push_back(ar[i][r]);
+            r--;
+            dir=2;
+             continue;
+        }
+        if(dir==2){
+            for(int i=r; i>=l; i--) res.push_back(ar[b][i]);
+            b--;
+            dir=3;
+             continue;
+        }
+        if(dir==3){
+            for(int i=b; i>=t;i--) res.push_back(ar[i][l]);
+            l++;
+            dir=0;
+             continue;
+        }
+    }
+    return res;
+}

competitive programming/leetcode/191. Number of 1 Bits.cpp

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+Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight).
+
+
+
+Example 1:
+
+Input: 00000000000000000000000000001011
+Output: 3
+Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.
+Example 2:
+
+Input: 00000000000000000000000010000000
+Output: 1
+Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.
+Example 3:
+
+Input: 11111111111111111111111111111101
+Output: 31
+Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.
+
+
+Note:
+
+Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
+In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 3 above the input represents the signed integer -3.
+
+
+Follow up:
+
+If this function is called many times, how would you optimize it?
+
+
+
+
+
+
+
+
+
+class Solution {
+public:
+    int hammingWeight(uint32_t n) {
+        int cnt=0;
+        int tmp=32;
+        while(tmp--){
+            bool a=n & 1;
+            if(a) cnt++;
+            n=n>>1;
+        }
+        return cnt;
+    }
+};
+
+
+
+
+
+
+class Solution {
+public:
+    int hammingWeight(uint32_t n) {
+        return __builtin_popcount(n);
+    }
+};

competitive programming/leetcode/2020-July-Challenge/Day-05-Hamming Distance.cpp

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+The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
+
+Given two integers x and y, calculate the Hamming distance.
+
+Note:
+0 ≤ x, y < 231.
+
+Example:
+
+Input: x = 1, y = 4
+
+Output: 2
+
+Explanation:
+1   (0 0 0 1)
+4   (0 1 0 0)
+       ↑   ↑
+
+The above arrows point to positions where the corresponding bits are different.
+
+
+
+
+
+
+
+
+
+
+class Solution {
+public:
+    int hammingDistance(int x, int y) {
+        int n=32, cnt=0;
+        while(n--){
+            bool a=x & 1;
+            bool b=y & 1;
+            if(a!=b) cnt++;
+            x=x>>1;
+            y=y>>1;
+        }
+        return cnt;
+    }
+};

competitive programming/leetcode/461. Hamming Distance.cpp

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+The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
+
+Given two integers x and y, calculate the Hamming distance.
+
+Note:
+0 ≤ x, y < 231.
+
+Example:
+
+Input: x = 1, y = 4
+
+Output: 2
+
+Explanation:
+1   (0 0 0 1)
+4   (0 1 0 0)
+       ↑   ↑
+
+The above arrows point to positions where the corresponding bits are different.
+
+
+
+
+
+
+
+
+
+
+class Solution {
+public:
+    int hammingDistance(int x, int y) {
+        int n=32, cnt=0;
+        while(n--){
+            bool a=x & 1;
+            bool b=y & 1;
+            if(a!=b) cnt++;
+            x=x>>1;
+            y=y>>1;
+        }
+        return cnt;
+    }
+};
+
+
+
+
+
+
+
+
+
+class Solution {
+public:
+    int hammingDistance(int x, int y) {
+        int res=x^y;
+        int cnt=0;
+        while(res>0){
+            cnt+=res & 1;
+            res>>=1;
+        }
+        return cnt;
+    }
+};