In this article, we need to find the number of segments or sub-arrays in a given sequence that are greater than a given number X.
We can only count overlapping segments once, two adjacent elements or segments should not be counted separately. So here is a basic example of the given problem −
Input : arr[ ] = { 9, 6, 7, 11, 5, 7, 8, 10, 3}, X = 7 Output : 3 Explanation : { 9 }, { 11 } and { 8, 10 } are the segments greater than 7 Input : arr[ ] = { 9, 6, 12, 2, 11, 14, 8, 14 }, X = 8 Output : 4 Explanation : { 9 }, { 12 }, { 11, 14 } and { 14 } are the segments greater than 8
In this problem we initialize the variables with 0state and start processing the given array, when an element greater than X is found, change the state to 1 and continue processing elements; when a number less than or equal to X is found, change the state back to 0, each time When the status changes to 1 and returns, count is increased by 1 to 0.
#include <bits/stdc++.h> using namespace std; int main (){ int a[] = { 9, 6, 12, 2, 11, 14, 8, 14 }; int n = sizeof (a) / sizeof (a[0]); int X = 8; int state = 0; int count = 0; // traverse the array for (int i = 0; i < n; i++){ // checking whether element is greater than X if (a[i] > X){ state = 1; } else{ // if flag is true if (state) count += 1; state = 0; } } // checking for the last segment if (state) count += 1; cout << "Number of segments where all elements are greater than X: " << count; return 0; }
Number of segments where all elements are greater than X: 4
In the above program, we use the state as a switch when a number greater than X is found Set it to 1, when we find a number greater than X we set it to 0 when we find a number less than or equal to Finally, print the results stored in the count.
In this article, we solved the problem of finding the number of segments where all elements are greater than X by applying a method of setting the state to 1 and 0 whenever a segment is found. We can write this program in any other programming language like C, Java, Python, etc.
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