# Minimum moves to reach from i to j in a cyclic string

Given a cyclic string str and two integers i and j, the task is to count the minimum number of steps required to move from str[i] to str[j]. A move is to reach any adjacent character in the string and the move is only counted if str[start] != start[end] where start is the starting index for the move and end is the ending (adjacent either on the left or on the right) index. Since, the given string is circular, str[0] and str[n – 1] are adjacent to each other.

Examples:

```Input: str = "SSNSS", i = 0, j = 3
Output: 0 From left to right : S -> S -> N -> S From right to left : S -> S -> S ```
```Input: str = "geeksforgeeks", i = 0, j = 3
Output: 2```

Approach:

• Starting from index i start moving in the right direction till index j and for every character visited, if the current character is not equal to the previous character then increment steps1 = steps1 + 1.
• Similarly, starting from i start moving in the left direction till index 0 and for every character visited, if the current character is not equal to the previous character then increment steps2 = steps2 + 1. Once the index 0 is visited, start traversing from index n – 1 to j and increment step2 if str[0] != str[n – 1].
• Print min(step1, step2) in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the count of steps` `// required to move from i to j` `int` `getSteps(string str, ``int` `i, ``int` `j, ``int` `n)` `{` `    ``// Starting from i + 1` `    ``int` `k = i + 1;`   `    ``// Count of steps` `    ``int` `steps = 0;`   `    ``// Current character` `    ``char` `ch = str[i];` `    ``while` `(k <= j) {`   `        ``// If current character is different from previous` `        ``if` `(str[k] != ch) {`   `            ``// Increment steps` `            ``steps++;`   `            ``// Update current character` `            ``ch = str[k];` `        ``}` `        ``k++;` `    ``}`   `    ``// Return total steps` `    ``return` `steps;` `}`   `// Function to return the minimum number of steps` `// required to reach j from i` `int` `getMinSteps(string str, ``int` `i, ``int` `j, ``int` `n)` `{`   `    ``// Swap the values so that i <= j` `    ``if` `(j < i) {` `        ``int` `temp = i;` `        ``i = j;` `        ``j = temp;` `    ``}`   `    ``// Steps to go from i to j (left to right)` `    ``int` `stepsToRight = getSteps(str, i, j, n);`   `    ``// While going from i to j (right to left)` `    ``// First go from i to 0` `    ``// then from (n - 1) to j` `    ``int` `stepsToLeft = getSteps(str, 0, i, n)` `                      ``+ getSteps(str, j, n - 1, n);`   `    ``// If first and last character is different` `    ``// then it'll add a step to stepsToLeft` `    ``if` `(str[0] != str[n - 1])` `        ``stepsToLeft++;`   `    ``// Return the minimum of two paths` `    ``return` `min(stepsToLeft, stepsToRight);` `}`   `// Driver code` `int` `main()` `{` `    ``string str = ``"SSNSS"``;` `    ``int` `n = str.length();` `    ``int` `i = 0, j = 3;` `    ``cout << getMinSteps(str, i, j, n);` `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach`   `class` `GFG` `{` `    ``// Function to return the count of steps` `    ``// required to move from i to j` `    ``static` `int` `getSteps(String str, ``int` `i, ``int` `j, ``int` `n)` `    ``{` `        ``// Starting from i + 1` `        ``int` `k = i + ``1``;` `    `  `        ``// Count of steps` `        ``int` `steps = ``0``;` `    `  `        ``// Current character` `        ``char` `ch = str.charAt(i);` `        ``while` `(k <= j) ` `        ``{` `    `  `            ``// If current character is different from previous` `            ``if` `(str.charAt(k) != ch)` `            ``{` `    `  `                ``// Increment steps` `                ``steps++;` `    `  `                ``// Update current character` `                ``ch = str.charAt(k);` `            ``}` `            ``k++;` `        ``}` `    `  `        ``// Return total steps` `        ``return` `steps;` `    ``}` `    `  `    ``// Function to return the minimum number of steps` `    ``// required to reach j from i` `    ``static` `int` `getMinSteps(String str, ``int` `i, ``int` `j, ``int` `n)` `    ``{` `    `  `        ``// Swap the values so that i <= j` `        ``if` `(j < i) ` `        ``{` `            ``int` `temp = i;` `            ``i = j;` `            ``j = temp;` `        ``}` `    `  `        ``// Steps to go from i to j (left to right)` `        ``int` `stepsToRight = getSteps(str, i, j, n);` `    `  `        ``// While going from i to j (right to left)` `        ``// First go from i to 0` `        ``// then from (n - 1) to j` `        ``int` `stepsToLeft = getSteps(str, ``0``, i, n)` `                        ``+ getSteps(str, j, n - ``1``, n);` `    `  `        ``// If first and last character is different` `        ``// then it'll add a step to stepsToLeft` `        ``if` `(str.charAt(``0``) != str.charAt(n - ``1``))` `            ``stepsToLeft++;` `    `  `        ``// Return the minimum of two paths` `        ``return` `Math.min(stepsToLeft, stepsToRight);` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main(String []args)` `    ``{` `        ``String str = ``"SSNSS"``;` `        ``int` `n = str.length();` `        ``int` `i = ``0``, j = ``3``;` `        ``System.out.println(getMinSteps(str, i, j, n));` `    ``}` `}`   `// This code is contributed by ihritik `

## Python3

 `# Python3 implementation of the approach`   `# Function to return the count of steps` `# required to move from i to j` `def` `getSteps( ``str``,  i, j, n) :`   `    ``# Starting from i + 1` `    ``k ``=` `i ``+` `1`   `    ``# Count of steps` `    ``steps ``=` `0`   `    ``# Current character` `    ``ch ``=` `str``[i]` `    ``while` `(k <``=` `j): `   `        ``# If current character is different from previous` `        ``if` `(``str``[k] !``=` `ch): `   `            ``# Increment steps` `            ``steps ``=` `steps ``+` `1`   `            ``# Update current character` `            ``ch ``=` `str``[k]` `        `  `        ``k ``=` `k ``+` `1` `    `    `    ``# Return total steps` `    ``return` `steps`     `# Function to return the minimum number of steps` `# required to reach j from i` `def` `getMinSteps( ``str``, i, j, n):`     `    ``# Swap the values so that i <= j` `    ``if` `(j < i):` `        ``temp ``=` `i` `        ``i ``=` `j` `        ``j ``=` `temp` `    `    `    ``# Steps to go from i to j (left to right)` `    ``stepsToRight ``=` `getSteps(``str``, i, j, n)`   `    ``# While going from i to j (right to left)` `    ``# First go from i to 0` `    ``# then from (n - 1) to j` `    ``stepsToLeft ``=` `getSteps(``str``, ``0``, i, n) ``+` `getSteps(``str``, j, n ``-` `1``, n)`   `    ``# If first and last character is different` `    ``# then it'll add a step to stepsToLeft` `    ``if` `(``str``[``0``] !``=` `str``[n ``-` `1``]):` `        ``stepsToLeft ``=` `stepsToLeft ``+` `1`   `    ``# Return the minimum of two paths` `    ``return` `min``(stepsToLeft, stepsToRight)`   `    `  `# Driver code`   `str` `=` `"SSNSS"` `n ``=` `len``(``str``)` `i ``=` `0` `j ``=` `3` `print``(getMinSteps(``str``, i, j, n))`   `# This code is contributed by ihritik `

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG` `{` `    ``// Function to return the count of steps` `    ``// required to move from i to j` `    ``static` `int` `getSteps(``string` `str, ``int` `i, ``int` `j, ``int` `n)` `    ``{` `        ``// Starting from i + 1` `        ``int` `k = i + 1;` `    `  `        ``// Count of steps` `        ``int` `steps = 0;` `    `  `        ``// Current character` `        ``char` `ch = str[i];` `        ``while` `(k <= j)` `        ``{` `    `  `            ``// If current character is different from previous` `            ``if` `(str[k] != ch)` `            ``{` `    `  `                ``// Increment steps` `                ``steps++;` `    `  `                ``// Update current character` `                ``ch = str[k];` `            ``}` `            ``k++;` `        ``}` `    `  `        ``// Return total steps` `        ``return` `steps;` `    ``}` `    `  `    ``// Function to return the minimum number of steps` `    ``// required to reach j from i` `    ``static` `int` `getMinSteps(``string` `str, ``int` `i, ``int` `j, ``int` `n)` `    ``{` `    `  `        ``// Swap the values so that i <= j` `        ``if` `(j < i) ` `        ``{` `            ``int` `temp = i;` `            ``i = j;` `            ``j = temp;` `        ``}` `    `  `        ``// Steps to go from i to j (left to right)` `        ``int` `stepsToRight = getSteps(str, i, j, n);` `    `  `        ``// While going from i to j (right to left)` `        ``// First go from i to 0` `        ``// then from (n - 1) to j` `        ``int` `stepsToLeft = getSteps(str, 0, i, n)` `                        ``+ getSteps(str, j, n - 1, n);` `    `  `        ``// If first and last character is different` `        ``// then it'll add a step to stepsToLeft` `        ``if` `(str[0] != str[n - 1])` `            ``stepsToLeft++;` `    `  `        ``// Return the minimum of two paths` `        ``return` `Math.Min(stepsToLeft, stepsToRight);` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``string` `str = ``"SSNSS"``;` `        ``int` `n = str.Length;` `        ``int` `i = 0, j = 3;` `        ``Console.WriteLine(getMinSteps(str, i, j, n));` `    ``}` `}`   `// This code is contributed by ihritik `

## PHP

 ``

## Javascript

 `// Javascript implementation of the approach`   `// Function to return the count of steps` `// required to move from i to j` `function` `getSteps(str, i, j, n)` `{` `    ``// Starting from i + 1` `    ``let k = i + 1;`   `    ``// Count of steps` `    ``let steps = 0;`   `    ``// Current character` `    ``let ch = str[i];` `    ``while` `(k <= j) {`   `        ``// If current character is different from previous` `        ``if` `(str[k] != ch) {`   `            ``// Increment steps` `            ``steps++;`   `            ``// Update current character` `            ``ch = str[k];` `        ``}` `        ``k++;` `    ``}`   `    ``// Return total steps` `    ``return` `steps;` `}`   `// Function to return the minimum number of steps` `// required to reach j from i` `function` `getMinSteps(str, i, j, n)` `{`   `    ``// Swap the values so that i <= j` `    ``if` `(j < i) {` `        ``let temp = i;` `        ``i = j;` `        ``j = temp;` `    ``}`   `    ``// Steps to go from i to j (left to right)` `    ``let stepsToRight = getSteps(str, i, j, n);`   `    ``// While going from i to j (right to left)` `    ``// First go from i to 0` `    ``// then from (n - 1) to j` `    ``let stepsToLeft = getSteps(str, 0, i, n)` `                      ``+ getSteps(str, j, n - 1, n);`   `    ``// If first and last character is different` `    ``// then it'll add a step to stepsToLeft` `    ``if` `(str[0] != str[n - 1])` `        ``stepsToLeft++;`   `    ``// Return the minimum of two paths` `    ``return` `Math.min(stepsToLeft, stepsToRight);` `}`   `// Driver code` `let str = ``"SSNSS"``;` `let n = str.length;` `let i = 0;` `let j = 3;` `console.log(getMinSteps(str, i, j, n));`   `// This code is contributed by Samim Hossain Mondal.`

Output

`0`

Complexity Analysis:

• Time Complexity: O(j)
• Auxiliary Space: O(1)

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