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Search a 2D Matrix in C++
Suppose we have write an efficient algorithm to searches for a value in one m x n matrix. This matrix has some properties like below −
- Each row is sorted from left to right
- The first number of each row is greater than the last integer of the previous row.
So if the matrix is like −
| 1 | 3 | 5 | 7 |
| 10 | 11 | 16 | 20 |
| 23 | 30 | 34 | 50 |
| 53 | 62 | 78 | 98 |
And if the target value is 16, then the output will be True.
Let us see the steps −
- n := number of rows, if n is 0, then return false, m := number of columns, if m = 0, then return false
- low := 0 and high := n – 1
- while low < high
- mid := low + (high – low + 1)/2
- if mat[mid, 0] <= target, then low := mid, otherwise high := mid – 1
- rlow := 0 and rhigh := m – 1 and ans := 0
- while rlow <= rhigh
- mid := rlow + (rhigh - rlow)/2
- if mat[low, mid] = target, then ans := 1, and break the loop
- otherwise when matrix[low, mid] < target, then rlow := mid + 1
- else rhigh := mid – 1
- return ans
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
lli n,m;
n = matrix.size();
if(!n)return false;
m = matrix[0].size();
if(!m)return false;
lli low = 0, high = n-1;
while(low<high){
lli mid = low + ( high - low +1)/2;
if(matrix[mid][0]<=target)low = mid;
else high = mid -1;
}
lli rlow = 0, rhigh = m-1;
lli ans = 0;
while(rlow<=rhigh){
lli mid = rlow+(rhigh - rlow)/2;
if(matrix[low][mid] == target){
ans =1;
break;
}else if(matrix[low][mid]<target)rlow=mid+1;
else rhigh= mid-1;
}
return ans;
}
};
main(){
Solution ob;
vector<vector<int>> v = {{1,3,5,7},{10,11,16,20},{23,30,34,50},{53,62,78,98}};
cout << ob.searchMatrix(v, 16);
}Input
[[1,3,5,7],[10,11,16,20],[23,30,34,50],[53,62,78,98]] 16
Output
1
Updated on: 04-May-2020
287 Views
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